4x2 + — b2 — 2ab a = 2x, b€¦ · RD Sharma Solutions Class 9 Algebraic Identities Ex 4.1 RD Sharma Solutions Class 9 Chapter 4 Ex 4.1 1) Evaluate each of the following using

RD Sharma Solutions Class 9 Algebraic Identities Ex 4.1

RD Sharma Solutions Class 9 Chapter 4 Ex 4.1

1) Evaluate each of the following using identities:

0)( 2 z - £ ) 2

Solution:

Given,

( 2 x - ± ) 2 = (2x)2 + (± ) 2 - 2 * 2 x * ±

(2x — - j ) 2 = 4x2 + — 4 [•.• (a — b)2 = a2 + b2 — 2ab ]

Where, a = 2x, b = ^

.-. (2a: — -i-)2 = 4a;2 + — 4

00 (2x+y) (2x-y)

Solution:

Given, (2x+y) (2x-y)

= (2a:)2 - (y ) 2 [•.■ (a + b)(a — b) = a 2 - b2]

= 4a:2 - y2

(2a: + y){2x - y) = 4a:2 - i/2

(Hi) (a 2b — ab2)2

Solution:

Given, (a 2b — ab2)2

= (a2b)2 + (ab2)2 — 2 * a 2b * ab2 [v (a — b)2 = a 2 + b2 — 2a6]

Where, a = a 2b, b = ab2

— a462 + b4a2 - 2a 3b3

(a 2b — ab2)2 — a4&2 + b4a2 — 2a363

(iv) (a-0.1) (a+0.1)

Solution:

Given, (a-0.1) (a+0.1)

= a 2 — (0.1)2 ['.• (a + b)(a — b ) = a 2 — b2]

Where, a = a and b = 0.1

= a 2 - 0.01

(o - 0.1)(o + 0.1) = o2 - 0.01

(v) (1.5a;2 - 0.3j/2)(1.5a:2 + 0.3y 2)

Solution:

Given, (1.5a;2 — 0.3y2)(1.5a:2 + 0.3y 2)

= (1.5a;2) 2 — (0.3 y 2)2 [•_• (a + b)(a — b) = a 2 — 62]

Where, a = 1.5a:2, b = 0.3y2

= 2.25a:4 - 0.09y4

.-. (1.5a:2 - 0.3y2)(1.5a:2 + 0.3i/2) = 2.25a:4 - 0.09y4

2) Evaluate each of the following using identities:

0) (399)2

Solution:

We have,

3992 = (400-1 )2

= (400)2+(1 )2 - 2x400x1 [ (a-b)2 = a2+ b2-2ab ]

Where, a = 400 and b = 1

= 160000 + 1 -8000

= 159201

Therefore, (399)2 = 159201.

(ii) (0.98f

Solution:

We have,

(0.98)2 = (1-0.02)2

= (1)2+(0.02)2- 2x1x0.02

= 1 + 0.0004 - 0.04 [ Where, a=1 and b=0.02 ]

= 1.0004-0.04

= 0.9604

Therefore, (0.98)2 = 0.9604

(Hi) 991x1009

Solution:

We have,

991x1009

= (1000-9) (1000+9)

= (1000)2 - (9)2 [ (a+b) (a-b) = a2 - b2 ]

= 1000000-81 [ Where a=1000 and b=9]

= 999919

Therefore, 991 xl 009 = 999919

(iv) 117x83

Solution:

We have,

117x83

= (100+17) (100-17)

= (100)2 - (1 i f [ (a+b) (a-b) = a2 - b2 ]

= 10000-289 [ Where a=100 and b=17 ]

= 9711

Therefore, 117x83 = 9711

3) Simplify each of the following:

0) 175x 175+2x 175x25+25x25

Solution:

We have,

175 x 175 +2 x 175 x 25 + 25 x 25 = (175)2 + 2 (175) (25) + (25)2

= (175+25)2 [ a2+ b2+2ab = (a+b)2 ]

= (200)2 [ Where a=175 and b=25 ]

= 40000

Therefore, 175 x 175 +2 x 175 x 25 + 25 x 25 = 40000.

00 322x322 - 2x322x22+22x22

Solution:

We have,

322 x 322 - 2 x 322 x 22 + 22 x 22

= (322-22)2 [ a2+ b2-2ab = (a-b)2 ]

= (300)2 [ Where a=322 and b=22 ]

= 90000

Therefore, 322 x 322 - 2 x 322 x 22 + 22 x 22= 90000.

OH) 0.76x0.76+2x0.76x0.24 + 0.24x0.24

Solution:

We have,

0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24

= (0.76+0.24)2 [ a2+ b2+2ab = (a+b)2 ]

= (1,00)2 [ Where a=0.76 and b=0.24 ]

= 1

Therefore, 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24 = 1.

0v) 7.83*7 .83-1 .17*1 .176.66

Solution:

We have.

7.83*7.83—1.17*1.176.66

_ (7 .83+ 1.17)(7.83—1.17) 6(66

_ (9.00)(6.66)(6.66)

[•.• (o - b)2 = ( a + b)(a — 6)]

. 7 .83*7 .83-1 .17*1 .17 _ q

6.66 y

4) I f x + ^ = 11, find the value o f x2 + 4

Solution:

We have, x + -j- = 11

Now, (a: + ± )2 = a:2 + ( )2 + 2 * x * j^ (x + ±)2 = x 2 + J _ + 2

=+(11)2 = x2 + 4 + 2 a: + ^ = 11]

=> 121 = x2 + 4 + 2 => a;2 + 4 = 119

X 25) I f x — — = —1, f in d the value o f x2 + 4 Solution:

We have, x — — = — 1X

Now, (x — |-)2 = a;2 + (-i-)2 — 2 * x *^(x-^)2 = x2 + 4 - 2^(-l)2 = a:2 + i - 2 [vx-± = -l]=>2 + 1 = a:2+ 4

X 2

6) I f x + = \/h, f in d the value o f x2 + 4 and x4 + 4 Solution:

We have,

(* + )2 = x2 + (i)2 + 2 * x * ±

^ ( X + ± f = X2 + ± + 2

=> (-s/5)2 = x2 + 4 + 2 [v x + = \/5]+►5 = a:2+ 4 + 2

X I^ * 2 + ^ = 3 ...(1)

Now, (a:2 + 4)2 = *4 + 4 + 2*a:2*4^ ( a : 2 + 4 ) 2 = x4 + 4 + 2

=>9 = x4 + 4 + 2 [••• a:2 + ^ = 3]

^ * 4 + V = 7

Hence, x2 + 4 = 3; x4 + 4 = 7-X £ X 4

7) I f x2 + 4 = 66, f in d the value o f x — ^

Solution:

We have,

(x-i)2 = x2 + (i)2-2*x*|^(x_i)2 = x2 + 4 - 2=> (x - |)2 = 66 - 2 [+ x2 + 4 = 66]

=► (* - | )2 = 64

= K * - i ) a = (± 8 )2

=► * - * = ± 8

8) I f x 2 + = 79, find the value o f x + ^

Solution:

We have,

( ! E + I ) 2 = a;2 + ( I ) 2 + 2 * a;* i

^ {x + 1 )2 = x 2 + ± + 2

^ ( x + |)2 = 79 + 2 [v*a + = 7 9 ]

= > (* + ± ) a = 81

=>(* + £ )a = (±9)a

=> * + i = ±9

9) if9x*+25f= 181 andxy=-6, find the value of 3x+5y.

Solution:

We have,

(3x + 5y)2 = (3x)2 + (5y)2 + 2*3x*5y

=>(3x + 5y)2 = 9x2 + 25y2 + 30xy

= 181+ 30(-6) [ Since, 9x2 + 25y2 = 181 and xy = -6 ]

=>(3x+5y)2 = 1

=>(3a; + by)2 = (±1)2

=>3* + by = ±1

10) lf2x+3y=8 and xy=Z find the value o f + 9f.

Solution:

We have,

(2x + 3y)2 = (2x)2 + (3y)2 + 2*2x*3y

=> (2x + 3y)2 = 4x2 + 9y2 + 12xy [Since, 2x + 3y = 8 and xy = 24 ]

=> (8)2 = 4x2 + 9y2 + 24

=> 64 - 24 = 4x2 + 9y2

=> 4x2 + 9y2 = 40

11) lf3x-7y=10 and xy=-1, find the value of Sx2+49y*.

Solution:

We have,

(2 - 7y)2 = (3x)2 + (-7y)2 - 2*3x*7y

=> (3x - 7y)2 = 9x2 + 49y2 - 42xy [Since, 3x - 7y = 10 and xy = -1 ]

=> (10)2 = 9x2 + 49y2 + 42

=>100- 42 = 9x2 + 49y2

=> 9x2 + 49y2 = 58

12) Simplify each of the following products:

0) - 36)(36 + \ a ) { \ a 2 + 9b2)

Solution:

(| a -3 6 )(3 6 + ± a )(| a 2 + 962)

=> K {« )2 - (3&)2)] \ { \a 2 + 9 ft2)] [•.• (a + b){a - b ) = a 2 - b2)}

^ [ \ a 2 -9t i2) ] [ \ a 2 + 9b?] [v (ab)2 = a2b2)\

= [(| «2)2 - (9&2)2] [••■ (a + 6 )(o - b ) = a2 - b2)\

= ^ a 4 -8 1 6 4

(\a - 36)(36+ f a ) ( {a 2 + 962) = ^ a 4 - 8164

0 0 { m + ± Y ( m - f )

Solution:

We have,

( m + y ) 3 ( m - y )

= ( ™ + f ) ( m + f ) ( m + f ) ( m - f )

= ( m + y ) 2 ( m 2 — ( y ) 2) [•.■ (a + 6)(a + 6) = (a + 6)2 an d (a + b)(a — 6) = a2 — 62]

= (m + y ) 2 K -( m + ? ) 3 ( m - ^ ) = ( m + ^ ) 2 ( m 2 - £ )

W ( f - f ) ( f - f ) - a;2 + 2a:

Solution:

We have,

(t - ! ) ( f - f ) - * 2 + 2*

= > - ( ! - ! ) (f - f ) - * 2 + 2*

=► - ( I - f ) 2 - X 2 + 2z [•.• (a - 6)(a - 6) = (a - 6)2]

^ - [ ( f )2 + ( ! ) 2 - 2(f ) (1 ) ] - * 2 + 2*

^ - ( ^ + T - f ) - ^ + 23:

^ - T - ^ + f + 2 x ~ i5x2 1 2 a ;____£

^ 4 ^ 5 25

• _ 2 \ /2 _ x\■ * ^2 5 M 5 2 * + 2z = ^ _4_

25

W ( * 2 + a; — 2) (a;2 — z + 2)

Solution:

( z 2 + x — 2) (z 2 — z + 2)

[(z ) 2 + (z - 2 ) ] [(z 2 - ( z + 2)]

=>(z2) 2 — (z — 2)2 [ (a - b) (a + b) = a2 - b2]

=>z4 — ( z 2 + 4 — 4 z) [y (a — 6)2 = a2 + 612 — 2a6]

=>z4 — z 2 + 4z — 4

(z2 + z — 2) (z2 — z + 2) = z4 — z2 + 4z — 4

(z3 — 3z — z) (z2 — 3z + 1)

Solution:

We have,

(z3 — 3z — z) (z2 — 3z + 1)

=>z(z2 — 3z — 1) (z2 — 3z + 1)

=>z[(z2 — 3z)2 — ( l )2] [•/ (a + 6)(a — 6) = a2 — 62]

=>z[(z2)2 + (—3z)2 — 2(3z)(z2) — 1]

=>z[z4 + 9z2 — 6z3 — 1]

=>z5 — 6z4 + 9z3 — z

(z 3 — 3z — z ) ( z 2 — 3z + 1) = z 5 — 6z4 + 9z3 — z

(vl) (2 z4 — 4 z2 + 1) (2 z4 — 4 z2 — 1)

Solution:

We have,

(2 z4 - 4 z2 + 1) (2 z4 - 4 z2 - 1)

=>[(2z4 - 4 z2) 2 - ( l ) 2] ['.• (a + 6)(0 - 6) = a2 - 9 ]

=>[(2£C4)2 + (4a:2)2 - 2(2a;4)(4a:2) - 1]

=>4a:8 — 16a:6 + 16a:4 — 1 [v (a — 6)2 = a2 + b2 — 2ab]

(2a:4 — 4a:2 + 1) (2a;4 — 4a:2 — 1) = 4a:8 — 16a:6 + 16a:4 — 1

13) Prove that a2 + b2 + c2 — ab — bc — ca is always non-negative for all values of a, b and c.

Solution:

We have,

a2 + b2 + C2 — ab — be — ca

Multiply and divide by '2'

= I [o2 + ft2 + c2 — ab — be — co]

= \ [2a2 + 2b2 + 2c2 - 2ab - 26c - 2co]

= -| [a2 + a2 + b2 + b2 + c2 + c2 — 2a6 — 26c — 2ca]

= [(a2 + 62 — 2a6) + (a2 + c2 — 2ca) + (62 + c2 — 26c)]

= 2 K « - ^)2 + (6 — c)2 + (c — a)2] [y (a — 6)2 = a2 + b2 — 2ab\

a2 + 62 + c2 — 06 — 6c — ca > 0

Hence, a2 + b2 + c2 — a6 — 6c — ca > 0 is always non-negative for all values of a, b and c.


RD Sharma Solutions Class 9 Chapter 4 Ex 4.2

Question 1: Write the following in the expand form:

(i) : (a + 26 + c)2

(ii) : (2a - 36 - c)2

(iii) : (-3a; + y + z)2

(iv) : (m + 2n — 5p)2

(v) : (2 + x - 2y f

(vi) : (a2 + 62 + c2) 2

(vii) : (ab + bc + ca) 2

(viii): ( f + f + f )2

(* > (£ + £ + ^ ) 2

(x): (x + 2y + 4z)2

W): ( 2 x - y + z)2

(xii): (~2x + 3 y + 2 z)2

Solution 1 (i):

We have,

(a + 2b c f — a2 + (26)2 + c2 2a(26) + 2ac + 2(26)c

[•.• (x + y + z f = x 2 + y 2 + z2 + 2 xy + 2yz + 2 xz]

(a + 26 + c)2 = a2 + 4b2 + c2 + 4ab + 2ac + 4be

Solution 1 (ii):

We have,

(2a - 36 - c)2 = [(2a) + (-36) + (-c)]2

(2a)2 + (—36)2 + ( - c )2 + 2(2a)(—36) + 2 (-36 )(-c ) + 2(2a)(-c)

[•.• (x + y + z)2 = x 2 + y 2 + z2 + 2 xy + 2 yz + 2 xz]

4a2 + 962 + c2 — 12a6 + 66c — 4ca

••• (2a-3b-c)A{2}=4xA{2}+9yA{2}+cA{2}-12ab+6bc-4ca

Solution 1 (iii):

We have,

(—3® + y + z)2 = [(-3®)2 + y2 + z2 + 2(-3®)j/ + 2yz + 2 (-3 x)z

[•.• (x + y + z)2 = x 2 + y 2 + z2 + 2 xy + 2 yz + 2 xz]

9®2 + y2 + z 2 — 6xy + 2 yz — 6 xz

(—3® + y + z)2 = 9®2 + j/2 + z2 — 6 xy + 2 xy — 6 xy

Solution 1 (iv):

We have,

(m + 2n — 5p)2 = m 2 + (2n )2 + ( — 5p)2 + 2m x 2n + (2 x 2n x — 5p) + 2m x — 5p

[v (® + y + z f = x2 + y2 + z2 + 2xy + 2yz + 2xz]

(m + 2n — 5p)2 = m 2 + 4n2 + 25p2 + 4mn — 20np — 10pm

Solution 1 (v):

We have,

(2 + x - 2y)2 = 22 + x 2 + (~2y)2 + 2(2)(®) + 2 (® )(-2 y ) + 2 (2 )(-2 y )

[y (® + y + z)2 = x 2 + y2 + z2 + 2xy + 2yz + 2xz]

= 4 + x 2 + 4 y2 + 4® — 4xy — 8 y

(2 + x — 2 y)2 = 4 + x 2 + 4 y2 + 4® — 4xy — 8 y

Solution 1 (vi):

We have,

(a2 + 62 + c2)2 = (a2)2 + (62)2 + (c2)2 + 2 a262 + 2b2 c2 + 2a2c2

[■.■ (® + y + z)2 = x 2 + y2 + z2 + 2 xy + 2 yz + 2 xz]

(a2 + 62 + c2)2 = a4 + 64 + c4 + 2a262 + 26V + 2c2a2

Solution 1 (vii):

We have,

(ab + be + ca)2 = (a&)2 + (6c)2 + (ca)2 + 2(a6)(6c) + 2(&c)(ca) + 2(a6)(ca)

[y (x + y + z)2 = x 2 + y 2 + z2 + 2xy + 2yz + 2xz]

= a2b2 + b2c2 + (?a2 + 2(ac)62 + 2(a6)(c)2 + 2(6c)(a)2

(ab + be + ca)2 = a262 + &2c2 + c2a2 + 2ac62 + 2abc2 + 2bca2

Solution 1 (viii):

We have.

( f + ! + 7 )2 = ( f )2 + ( ! ) 2 + ( I )2 + 2 f 7 + 2f 7 + 2^ f['.• (a: + y + z)2 = a;2 + y2 + z2 + 2xy + 2 yz + 2 xz\

■ • • ( f + f + i )2 = ( S ) + ( S ) + ( ^ ) + 2 f + 2 f + 2fSolution 1 (ix):

We have,

( e + = + ^ ) 2 = ( £ ) 2 + ( i r ) 2 + ( ^ ) 2 + 2 ( £ ) ( £ ) + 2( £ ) ( £ ) + 2( e ) ( 3 )

[■.■ (a-, + y + z)2 = x 2 + y 2 + z2 + 2xy + 2 yz + 2xz]( «_ + A _£.)2 = (_aL\ , ( J L ) + (_*!_) _|— 2— |— 2— |— 2_t be ' co ~ ab > 'P62 1 ' Kcla2 ' ' 'aW ' ~ a? ' & ' c2Solution 1 (x):

We have,

(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + 2x x 2y + 2 x 2y x 4z + 2x x Az

[•.• (x + y + z)2 = x 2 + y 2 + z2 + 2xy + 2 yz + 2 xz]

(a; + 2 y + 4z)2 = x2 + 4 y2 + 16 z 2 + 4 xy + 16 yz + 8 xz

Solution 1 (xi):

We have,

(2 x - y + z)2 = (2a:)2 + { -y )2 + (z)2 + 2(2 x ){ -y ) + 2(-y)(z) + 2(2 x)(z)

[■.• (a:+ y + z)2 = x 2 + y2 + z2 + 2 xy + 2 yz + 2xz]

(2x - y + z)2 = Ax2 + y 2 + z 2 — 4a:y — 2yz + 4xz

Solution 1 (xii):

We have,

(-2a : + 3 y + 2 z)2 = (-2 a :)2 + (3y)2 + (2*)2 + 2(— 2as)(3y) + 2(3y)(2z) + 2 (-2® )(2z)

[■.■ (x + y + z)2 = x 2 + y 2 + z2 + 2 xy + 2 yz + 2 xz]

( — 4® + 6 y + 4 z)2 = 4®2 + 9 y2 + 4 z 2 — 12xy + 12 yz — 8 xz

Question 2: Use algebraic identities to expand the following algebraic equations.

Q2.1:(a + b + c)2 + (a — b + c)2

Ans: We have,

(a + b + c)2 + (a — b + c)2 =(a2 + 62 + c2 + 2ab + 2bc + 2 co) + (a2 + (~b)2 + c2 — 2 ab — 26c + 2co)

[v (® + y + z)2 = x 2 + y2 + z2 + 2xy + 2yz + 2xz]

= 2o2 + 262 + 2c2 + 4ca

(a + b + c)2 + (a —b + c)2 = 2 a2 + 2b2 + 2c2 + 4ca

Q2.2: (a + b + c)2 — (a —b + c)2

Ans: We have,

(a + b + c)2 — (a —b + c)2 = (a2 + b2 + C2 + 2ab + 26c + 2co)

- (a2 + ( — 6)2 + c2 - 2a6 - 26c + 2ca)

['.• (x + y + z)2 = x 2 + y 2 + z2 + 2 xy + 2 yz + 2 xz]

= a2 + ti2 + c2 + 2ab + 26c + 2ca — a2 — 62 — c2 + 2a6 + 26c — 2co)

= 4o6 + 46c

(a + 6 + c)2 — (a — 6 + c)2 = 4a6 + 46c

Q2.3: (a + b + c)2 + (a + 6 — c)2 + (a + b — c)2

Ans: We have.

(a + b + c)2 + (a + b — c)2 + (a + b — c)2 = a2 + b2 + <? + 2ab + 26c + 2ca

+ (a2 + 62 + (z)2 — 26c — 2o6 + 2ca) + (a2 + 62 + c2 — 2ca — 26c + 2ab)

[•.• (x + y + z)2 = x 2 + y 2 + z2 + 2 xy + 2 yz + 2 xz]

— 3 a2 + 362 + 3c2 + 2a6 + 26c + 2ca — 26c — 2a6 — 2ca — 26c + 2ab

= 3®2 + 3 y 2 + 3 z 2 + 2ab — 26c + 2ca

(a+ b + c)2 + (a + b — c)2 + (a — b + c)2 = 3a2 + 362 + 3c2 + 2 ab — 26c + 2ca

(a + 6 + c)2 + (a + 6 — c)2 + (a — 6 + c)2 = 3(a2 + 62 + c2) + 2(ab — bc + ca)

Q2.4: (2x + p — c)2 — (2x — p + c)2

Arts: We have,

(2® + p - c)2 — (2x - p + c)2 = [2a;2 + p2 + ( - c ) 2 + 2(2®)p + 2 p (-c ) + 2(2®)(— c)]

- [4a;2 + ( - p ) 2 + c2 + 2(2®)(—p) + 2 c (-p ) + 2(2®)c]

[y (® + p + z )2 = ®2 + p2 + z2 + 2xy + 2yz + 2xz]

(2x + p — c)2 — (2a; — p + c)2 = [4®2 + p2 + c2 + 4®p — 2pc — 4®c]

- [4®2 + p2 + c2 — 4®p — 2pc + 4®c]

Opening the bracket,

(2x + p — c)2 — (2x — p + c)2 = 4a;2 + p2 + c2 + 4®p — 2pc — 4c® — 4®2 — p2 — c2 + 4®p

+ 2pc — 4c®]

(2® + p — c)2 — (2® — p + c)2 = 8®p — 8®c

=8 ®(p — c)

Hence, (2® + p — c)2 — (2® — p + c)2 = 8®(p — c)

Q2.5:(x2 + y 2 + ( - z ) 2) - (®2 - y2 + z2) 2

Ans: We have,

(®2 + p 2 + ( - z ) 2) 2 - ( ® 2( - p ) 2 + z 2) 2

[®4 + p4 + ( - z ) 4 + 2x2y 2 + 2 y2( - z ) 2 + 2®2( - z ) 2]

- [®4 + ( - p ) 4 + z4 - 2®2p2 - 2p2z2 + 2®2z 2]

['.• (® + p + z )2 = ®2 + p2 + z2 + 2®p + 2 pz + 2®z]

Taking the negative sign inside,

[®4 + p4 + ( - z ) 4 + 2®2p2 + 2 p2( - z ) 2 + 2®2( - z) 2]

- [®4 + ( - p ) 4 + z4 - 2®2p2 - 2p2z2 + 2®2z 2]

= 4®2p2-4 z 2®2

Hence, (®2 + p2 + ( — z )2) 2 — (®2( —p)2 + z2)2 = 4®2p2-4 z 2®2

Q3: lfa+b+c=0 and a2 + 62 + c2 = 16, find the value o f ab+bc+ca:

Ans: We know that,

[y (a + 6 + c)2 = a2 + 62 + c2 + 2a6 + 26c + 2ca]

(0)2=16+2(ab+bc+ca)2(ab+bc+ca)=-16 ab+bc+ca=-8

Hence, value of required express ab+bc+ca =-8

Q4: If a2 + br2 + c2 = 16 and ab+bc+ca=10, find the value ofa+bn:?

Ans: We know that,

(a + 6 + c)2 = a2 + 62 + c2 + 2(a6 + 6c + ca)(® + p + z )2 = 16 + 2(10)(® + p + z )2 = 36 (® + p + z) = i/36 (® + p + z) = ±6

Hence, value of required expression I; (a + 6 + c) = ±8

Q5: lfa+b+c=9 and ab+bc+ca=23, find value o f a2 + 62 + c2

Ans:\Ne know that,

(a + 6 + c)2 = a2 + 62 + c2 + 2(a6 + be + ca)92 = a2 + 62 + c2 + 2(23)81 = a2 + 62 + c2 + 46 a2 + 62 + c2 = 81 - 46 a2 + 62 + c2 = 35

Hence, value of required expression a2 + b2 + c2 = 35

Q6: Find the value of the equation: Ax2 + y2 + 25 z2 + 4 xy — 10 yz — 20 zx when x=4,y=3^=2

Ans: 4x2 + y2 + 25 z2 + 4xy — 10 yz — 20 zx

(2a:)2 + y2 + ( - 5 z)2 + 2(2 x)(y) + 2(y)(— 5z) + 2(—5z)(2x)(2x + y — 5 z)2 (2(4) + 3 — 5(2))2 (8 + 3 - 10)2 ( l ) 2 1

Hence value of the equation is equals to 1

Q7: Simplify each of the following expressions:

Q7.1:(x + y + z)2 + (x + f + f ) 2 - ( f + f + f ) 2

Ans: Expanding, we get

= [a;2 + y2+ z 2 + 2 xy + 2 yz + 2za;] + [a:2 + ^ + T + 2x| + 2x + f ]

[v (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

= x2 + y2 + z2 + 2xy + 2yz + 2zx + a;2 + ^ ^ + 2a:| + ^ ^ - £_ HI _ ££

6 4

Rearranging coefficients,

8a;2—x2 . 36y2+9y2—4y2 144z2+16z2-9z2 . 6xy+3xy-xy 13 yz 29 xz4 ' 36 + 144 "l" 3 + 6 + 12

_ 7if_ , 41y 151z2 , 8xy“ 4 + 36 + 144 + 3

132/3 29zx6 12

(a, + „ + 2)» + (, + | + i ) . _ ( f + I + z). = ^ + a . + ^ + ^ + lfc + ^Q 7.2: (x + y — 2 z)2 — x 2 — y2 — 3 z 2 + 4 xy

Ans: (x + y — 2 z)2 — x 2 —y2 — 3 z2 + 4a:y

= [a:2 + y2 + 4z 2 + 2xy + 2y(-2z) + 2a(-2c)] - x 2 - y 2 - 3z2 + 4xy

— z 2 + 6 xy — 4 yz — 4zx

(x + y — 2 z)2 — x 2 — y2 — 3 z2 + 4 xy = z2 + 6xy — 4 yz — 4 zx

Q 7.3: [x2 - x + l}2 - [x2 + x + l}2

Ans: [x2 — x + l ]2 — [x2 + x + l ]2

= (a:2) 2 + ( -a : )2 + l 2 + 2(a:2) ( -a :) + 2 (-a s )(l) + 2a:2)

— \{x2)2 + a:2 + 1 + 2x 2x + 2a:(l) + 2a:2(l)]

[■.• (x + y + z)2 = x 2 + y2 + z2 + 2 xy + 2 yz + 2xz]

= x4 + y2 + 1 - 2a:3 - 2a: + 2a:2 - a:2 - a:4 - 1 - 2a:3 - 2a: - 2a:2

= -4a:3 - 4a:

= —4a:(a:2 + 1)

Hence simplified equation =[x2 — x + l ]2 — [a:2 + x + l ]2 = — 4a:(a:2 + 1)

RD Sharm a Solutions Class 9 A lgebraic Identities Ex 4.3

RD Sharma Solutions Class 9 Chapter 4 Ex 4.3 Q1. Find the cube o f each of the following binomial expression

w ( 7 + ! )

(b) ( f - J r )(c) ( 2 x + ■§)

< « 0 ( 4 - £ )

S o l:

( a ) ( i + f ) ) 3

Given, ( I + f ) ) 3

The above equation is in the fo rm o f (a + b )3 = a3 + b3 + 3ab(a + b)

We know tha t, a = — , b = -|x 3

By using (a + b)3 fo rm u la

( i + f ) ) 3= ( i ) 3+ ( f ) 3+3 ( i ) ( f ) ( i + ! )

= ^ + £ + 3 * M ( M )

= ^ + £ + f ( i r + f )

+ ( ! * i ) +'.T .T'27

' x 3 '

= j _ + £. + jl + yL)x 3 27 X2 3x >

Hence,

G + l w - i + t + i + £ )

( b ) ( ( f - J r ) ) 3

Given, ( ( | - ^ ) ) 3

The above equation is in th e fo rm o f (a - b )3 = a3 - b3 - 3ab(a - b)

We know tha t, a = | - , b = -4-

By using (a - b)3 fo rm u la

( d - ^ ) ) 3= ( | ) 3- ( ^ ) 3- 3 ( l ) ( ^ ) ( l - | = 2 Z . - J . - 3 * 2 * A r l - A t

x3 X 6 X x2 ' x X 2

27 _ _8_ _ 1 8 / 3 _ _2_\x3 x6 x3 ' x x2

27 _ ± - ( I® * 2 1 + f i i * AX 3 X 6 ' X 3 ® ' ' X 3 X 2

27 _ _ 8_ _ 54 + 36

27 _ _8_ _ 54 36Hence, ( ( f - J r ) ) 3

(c) (2a; + f - ) 3

Given, (2a; + | ) 3

The above equation is in th e fo rm o f (a + b )3 = a3 + b3 + 3ab(a + b)q

We know tha t, a = 2 x , b = —

By using (a + b)3 fo rm u la

= 8x 3 + ^ - + ^ ( 2x + | )

= 8x 3 + 4 - + ^ ( 2 x + | )x 3 X ' X /

= 8x3 + ^ +X 3

( 1 8* 2 x ) + ( 1 8 * | )

= 8x3 + ^ + 36x

Hence,

The cube o f (2a; + - | ) 3 = 8x3 + ^ + 36x

( d ) ( 4 - ^ ) 3

Given, ( 4 - ^ ) 3

The above equation is in th e fo rm o f (a - b )3 = a3 - b3 - 3ab(a - b)

We know th a t ,a = 4 , b = ^

By using (a - b)3 fo rm u la

( 4 - ^ ) 3 = 4 3 - ( ^ ) 3 - 3 ( 4 ) ( - L ) ( 4 - 3 L )

- ^ “ 27x3 ~ 3 x ( 4 “ 3 ^ )

= 64-----------— (4 — —)W 27a;3 * ^ 3 * '

= 64 - 1 - (A * 41 + (A. * J - l27a:3 3a: 3a: 3a:'

= 6 4 ------1___ — + ( —

H e n ce ,

T h e c u b e o f(4 --L )3 = 6 4 - ^ - ^ + ( £ )

Q2. Simplify each of the following

(a) (x + 3) 3 + ( x - 3) 3

( b ) ( f + ! ) 3 - ( f - ! ) 3

(c) (* + f ) 3 + (*“ f ) 3(d) (2x - 5y) 3 - (2x + 5y)3

S o l:

(a) (x + 3 )3 + ( x - 3 )3

The above equation is in th e fo rm o f a3 + b3 = (a + b )(a2 + b2 - ab)

We know tha t, a = (x + 3 ) , b = (x - 3)

By using (a3 + b3) fo rm u la

= (x + 3 + x - 3 )[ (x + 3 )3 + (x - 3 )3 - (x + 3)(x - 3)]

= 2x[(x2 + 3 2 + 2 *x*3 ) + (x2 + 3 2 - 2 *x*3 ) - (x2 - 32)]

= 2x[(x2 + 9 + 6x) + (x2 + 9 - 6x) - x 2 + 9]

= 2x[(x2 + 9 + 6x + x2 - 9 - 6x - x 2 + 9)]

= 2x ( x2 + 27)

= 2x3 + 54x

Hence, the resu lt o f (x + 3 )3 + ( x - 3 )3 is 2x3 + 54x

(b)(f + f ) 3- ( f - | ) 3The above equation is in th e fo rm o f a3 - b3 = (a - b )(a2 + b2 + ab)

We know tha t, a = ( | + | ) 3 ,b = ( f — f ) 3

By using (a3 - b3) fo rm u la

= K(f + -f )3 - ( ( f - f )3 )][((# + f )3)2 ((f - f )3)2 - ( (f + f )3 ) ( ( f - f )3)( f + f - f + ! ) [ ( ( l )2 + ( f ) 2 + (3 r)) + ( ( f ) 2 + ( ! ) 2- ( ^ ) ) + ( ( f )2- ( f ) 2)]2 x y x

yLu* + y~ +3 LV 4 T 9 T

Zn\\22 x y \ _ i_ ( £ _ i y _ _ Zxy \ i V_

f 9 6 ' 4 9

_ r s 2 ■ V 2 ■ . x * . y ^ _ _ 2 x y _ y 2 n

“ 3 U “h 9 “h 6 “, _ 4 “h 9 6 “l~ 4 9 J

- ^ r £ l \ y2 \ x2 \ * i i- 3 L 4 9 4 ~■ 4 J

- ^ [ 3 * 1 i t . ]3 L 4 “l" 9 J

_ s 2y , 2y3 2 27

Hence, the resu lt o f ( f + | ) 3 - ( | - | ) 3 = ^ + ^ -

(c) ( * + l ) 3 + ( * - l ) 3The above equation is in th e fo rm o f a3 + b3 = (a + b )(a2 + b2 - ab)

We know tha t, a = (a: + - | ) 3 , b = (x- 1 -)3

By using (a3 + b3) fo rm u la

= ( * + ! + * - ! ) [ ( * + 1 ) 2 + ( * - i ) 2- ( ( * + ! ) ( * - ! ) ) ]

= (2 a :)[(a ;2 + £ + f ) + ( * 2 + H * 2 - £ )

= ^ [ ( z 2 + 4 r + f + x 2 + + ± )

= (2x) [ ^ + ± + ± + ± )

= {2x)[{x* + ^ )

= 2x3 + f

Hence, the resu lt o f (x + | - ) 3 + (x - | - ) 3 = (2a:) [(a:2 + ^ | )

(d) (2x - 5y)3 - (2x + 5y)3

Given, (2x - 5y)3 - (2x + 5y)3

The above equation is in th e fo rm o f a3 - b3 = (a - b )(a2 + b2 + ab)

We know tha t, a = (2x - 5 y ) , b = (2x + 5y)

By using (a3 - b3) fo rm u la

= (2x - 5y - 2x - 5y)[(2x - 5y)2 + (2x + 5y)2 + ((2x - 5y) * (2x + 5y» ]

= (-10y)[(4x2 + 25y2 - 20xy) + (4x2 + 25y2 + 20xy) + 4x2 - 25y2]

= (-10y)[ 4x2 + 25y2 - 20xy + 4 x 2 + 25y2 + 20xy + 4x2 - 25y2]

= (-10y)[4x2 + 4x2 + 4 x2 + 25y2]

= (-10y)[12x2 + 25y2}

= -120x2y - 250y3

Hence, the resu lt o f (2x - 5y)3 - (2x + 5y)3 = -120x2y - 250y3

Q3. If a + b = 10 and ab = 21, Find the value of a3 + b3

S o l:

Given,

a + b = 10, ab = 21

w e know tha t, (a + b )3 = a3 + b3 + 3ab(a + b) ------- 1

s u b s titu te a + b = 1 0 , ab = 21 in eq 1

=> (10 )3 = a3 + b3 + 3(21 )(1 0)

=> 1000 = a3 + b3 + 630

=> 1000 - 630 = a3 + b3

=> 370 = a3 + b3

Hence, the value o f a 3 + b3 = 370

Q4. If a - b = 4 and ab = 21, Find the value of a3 - b3

S o l:

Given,

a - b = 4 , ab = 21

w e know tha t, (a - b )3 = a3 - b3 - 3ab(a - b) ------- 1

s u b s titu te a - b = 4 , ab = 21 in eq 1

=> (4 )3 = a3 - b3 - 3(21 )(4)

=> 64 = a3 - b3 - 252

=> 64 + 252 = a3 - b3

= > 3 1 6 = a3 - b3

Hence, the value o f a 3 - b3 = 316

Q5. If (a; + = 5 , Find the value o f x 3 + ^

S o l:

Given, (x + j ) = 5

We know tha t, (a + b )3 = a3 + b3 + 3ab(a + b)

S ubstitu te (x + ^ -) = 5 in eq 1

(x + i ) 3 = x 3 + ± + 3 (x * i ) ( x + i )

53 = x3 + ^ + 3(x * I ) ( x + | )

125 = x3 + J3 + 3 ( x + i )

— 1

125 = x3 + ^ + 3 (5 )

125 = x 3 + ^ + 1 5X s

1 2 5 - 15 = x3 + \X A

x 3 + 4 r = 110®3

hence, th e resu lt is x 3 + -V = 1 1 0X s

Q6 . If (x - = 7 , Find the value of x3-

S o l:

Given, If (x- ^ ) = 7

We know tha t, (a - b )3 = a3 - b3 - 3ab(a - b) ------- 1

S ubstitu te (a :- = 7 in eq 1

73 = x3 - ^ a - 3 ( x - | )

343 = x3 - ^ - ( 3 * 7 )

343 = x 3 - - 21X3

343 + 21 = x3 -r-X 3

x 3 - 4 = 3 6 4X 3

hence, th e resu lt is x 3 - = 364X 3

Q7. If ( x - = 5 , Find the value o f a;3- ^

S o l:

Given, If ( x - j ) = 5

We know tha t, (a - b )3 = a3 - b3 - 3ab(a - b) ------- 1

S ubstitu te ( x - - ) = 5 in eq 1

( * - ¥ ) 3 = x3 - : ? - 3 ( x * ¥ ) ( x - f )

53 = x3 - ^ - 3 ( x - i )

1 2 5 = x 3 - ^ - ( 3 * 5 )

125 = x 3 - \ - 15Xs

125 + 15 = x3 - -VX 6

x 3 - = 140X 3

hence, th e resu lt is x 3 - -V = 140X 3

Q8 . If (x2 + = 5 1 , Find the value of x3- ^

S o l:

Given, (x2 + ± ) = 51

We know th a t , ( x - y )2 = x2 + y2 - 2xy ------- 1

S ubstitu te (a:2 + ^ - ) = 51 in eq 1

( x - ^ ) 2 = x 2 + ^ - 2 * x * i

(x - ¥ ) 2 = x 2 + ^ - 2

( x - i )2 = 5 1 - 2

(x - ^ ) 2 = 4 9

( x - I ) = V 3 9

( * - £ > = * 7

We need to find a:3-

So, a3 - b3 = (a - b )(a2 + b2 + ab)

a;3“ ^ = (x _ 7 ) ( xA{2>+ ^ + (x * ¥ )

We know th a t ,

( x - | ) = 7 and (a:2 + = 51

* 3- £ = 7 (5 1 + 1 )

* 3— ^ = 7 ( 5 2 )

a:3- ^ = 364

Hence, the value o f x3- = 364

Q9. If (a;2 + = 9 8 , Find the value o f a;3 + ^

S o l:

Given, (a;2 + ^ ) = 98

We know th a t , ( x + y )2 = x2 + y2 + 2xy ------- 1

S ubstitu te (a;2 + p - ) = 98 in eq 1

(x + - ) 2 = x2 + - V + 2 * x * -

(X + - ) 2 =X 2 + + 2

(x + ^ ) 2 = 9 8 + 2

(x + i ) 2 = 1 0 0

( x + i ) = v / l0 0

( x + I ) =±10

We need to find a;3 + -V

So, a3 + b3 = (a + b )(a2 + b2 - ab)

* 3 + ^ - = ( x + | ) ( x 2 + ^ - - ( x * ^ )

We know that,

( x + ^ ) = 10 and (a;2 + ^ j ) = 98

a:3 + ^ - = 1 0 ( 9 8 - 1 )

a:3 + ^ = 1 0 ( 9 7 )

a;3 + -V = 970

Hence, the value o f a:3 + = 970ar

Q10. If 2x + 3y = 13 and xy = 6, Find the value of 8X3 + 2 7 ^

S o l:

G ive n , 2x + 3y = 1 3 , xy = 6

We know that,

(2x + 3y)3 = 132

=> 8x3 + 27y3 + 3 (2x)(3y)(2x + 3y) = 2197

=> 8x3 + 27y3 + 18xy(2x + 3y) = 2197

S ubstitu te 2x + 3y = 13, xy = 6

=> 8x3 + 27y3 + 18 (6 )(13) = 2197

=> 8x3 + 27y3 + 1404 = 2197

=> 8x3 + 27y3 = 2 1 9 7 - 1404

=> 8x3 + 27y3 = 793

Hence, the value o f 8x3 + Hy3 = 793

Q11. If 3x — 2y =11 and xy= 12, Find the value of 27x3 - 8 y 3

S o l:

Given, 3x - 2y = 11 , x y = 12

We know th a t (a - b)3 = a3 - b3 - 3ab(a + b)

(3x - 2y)3 = 113

= > 27x3 - 8y3 - ( 1 8 * 1 2 * 1 1 ) = 1331

=> 27x3 - 8y3 - 2376 =1331

=> 27x3 - 8y3 = 1331 + 2 3 7 6

=> 27x3 - 8y3 = 3707

Hence, the value o f 27x3 - 8y3 = 3707

Q12. If as4 + ( ^ j ) = 119, Find the value o f z 3- ( ^ )

Sol:

G iv e n ,*4 + ( ^ j ) = 119 ------- 1

We know th a t (x + y )2 = x2 + y2 + 2xy

S ubstitu te * 4 + ( ^ - ) = 119 in eq 1

( * 2 + ( ^ ) ) 2 = x4 + ( £ ) + P**2* j l )

= x4 + ( ^ ) + 2

= 119 + 2

= 121

( * 2 + ( ^ ) ) 2 =121

x 2 + ( ^ ) = ^ 1 2 1

a:2 + ( J _ ) = +11

Now, find (x - -i-)

We know th a t (x - y )2 = x2 + y2 - 2xy

(x - ^ ) 2 = x2+ ^ - ( 2 * x4

= x2 + A “ 2X 1

= 11-2

= 9

( * - £ ) = > /§

= ±3

We need to find * 3- ( ^ - )

We know tha t, a3 - b3 = (a - b )(a2 + b2 - ab)

* 3- ( ^ ) = ( x - 7 X * 2 + ( ^ ) + x 4

Here, x2 + (■^■) = 11 and (x - = 3

z 3- ( ^ ) = 3 (1 1 + 1 )

= 3(12)

= 36

Hence, the value o f x3- (■^■) = 36

Q13. Evaluate each of the following

(a) (103)3

(b ) (98)3

(c) (9.9) 3

(d) (10.4)a

(e) (598)3

(f)(9 9 ) 3

S o l:

Given,

(a) (103 )3

w e know th a t (a + b )3 = a3 + b3 + 3ab(a + b)

=> (103)3 can be w ritte n as (100 + 3 )3

H e re , a = 100 and b = 3

(103)3 = (100 + 3 )3

= (100 )3 + (3 )3 + 3 (100 )(3 )(100 + 3)

= 1 0 0 0 0 0 0 + 2 7 + (900*103)

= 1000000 + 27 + 92700

= 1092727

The va lue o f (10 3 )3 = 1092727

(b) (98 )3

w e know th a t (a - b )3 = a3 - b3 - 3ab(a - b)

=> (98 )3 can be w ritte n as (100 - 2 )3

H e re , a = 100 and b = 2

(98 )3 = ( 1 0 0 - 2 ) 3

= (100 )3 - (2 )3 - 3 (100 )(2 )(100 - 2)

= 1 0 0 0 0 0 0 - 8 - ( 6 0 0 * 1 0 2 )

= 1 0 0 0 0 0 0 - 8 - 58800

= 941192

The va lue o f (98 )3 = 941192

(c) (9.9)3


=> (9 .9)3 can be w ritte n as (10 - 0.1 ) 3

H e re , a = 10 and b = 0.1

(9 .9)3 = ( 1 0 - 0 .1 )3

= (10 )3 - (0.1 ) 3 - 3 (10)(0.1 )(1 0 - 0.1)

= 1000 - 0.001 - (3*9.9)

= 1 0 0 0 -0 .0 0 1 - 2 9 .7

= 1 0 0 0 -2 9 .7 0 1

= 970.299

The va lue o f (9 .9 )3 = 970.299

(d) (10 .4 )3

w e know th a t (a + b )3 = a3 + b3 + 3ab(a + b)

=> (10 .4 )3 can be w ritte n as (10 + 0 .4 )3

H e re , a = 10 and b = 0.4

(10 .4 )3 = (10 + 0 .4 )3

= (10 )3 + (0 .4 )3 + 3 (10)(0 .4 )(10 + 0.4)

= 1 0 0 0 + 0 .0 6 4 + (12*10.4)

= 1000 + 0 .064 + 124.8

= 1000 + 124.864

= 1124.864

The va lue o f (10 .4 )3 = 1124.864

(e) (598)3


=> (598)3 can be w ritte n as (600 - 2 )3

H e re , a = 600 and b = 2

(598)3 = (600 - 2) 3

= (600)3 - (2 )3 - 3 (600)(2 )(600 - 2)

= 216000000 - 8 - (3600*598)

= 216000000 - 8 - 2152800

= 216000000 - 2152808

= 213847192

The va lue o f (598 )3 = 213847192

( f) (9 9 )3


=> (99 )3 can be w ritte n as ( 1 0 0 - 1 ) 3

H e re , a = 100 and b = 1

(99 )3 = ( 1 0 0 - I ) 3

= (100 )3 - ( I ) 3 - 3 (1 00 )(1 )(100 - 1)

= 1000000 - 1 - (300*99)

= 1 0 0 0 0 0 0 - 1 -2 9 7 0 0

= 1 0 0 0 0 0 0 -2 9 7 0 1

= 970299

The va lue o f (99 )3 = 970299

Q14. Evaluate each of the following

(a) 1113 - 893

(b) 463 + 343

(c) 1043 + 963

(d) 933 - 1073

Sol:

Given,

(a) 1113 - 893

the above equation can be w ritte n as (100 + 11 ) 3 - (100 - 11 ) 3

w e know tha t, (a + b )3 - (a - b )3 = 2 [b3 + 3ab2]

here, a= 100 b = 11

(100 + 11 ) 3 - (100 - 11 )3 = 2 [113 + 3 (100 )2(1 1)]

= 2 [1331 + 3 3 0 0 0 0 ]

= 2(331331]

= 662662

The va lue o f 1113 - 89 3 = 662662

(b) 4 6 3 + 343

the above equation can be w ritte n as (40 + 6 )3 + (40 - 6 )3

w e know tha t, (a + b )3 + (a - b )3 = 2 [a3 + 3ab2]

here, a= 4 0 , b = 4

(40 + 6 )3 + (40 - 6 )3 = 2 [403 + 3 (6 )2(40)]

= 2(64000 + 4320]

= 2(68320]

= 1366340

The va lue o f 46 3 + 343 = 1366340

(c) 1043 + 963

the above equation can be w ritte n as (100 + 4 )3 + ( 1 0 0 - 4 )3

w e know tha t, (a + b )3 + (a - b )3 = 2 [a3 + 3ab2]

here, a= 100 b = 4

(100 + 4 )3 - (100 - 4 )3 = 2 [1003 + 3 (4 )2(100)]

= 2(1000000 + 4800]

= 2(1004800]

= 2009600

The va lue o f 1043 + 96 3 = 2009600

(a) 933 - 107 3

the above equation can be w ritte n as (100 - 7 )3 - (100 + 7 )3

w e know tha t, (a - b )3 - (a + b )3 = -2 [b3 + 3ba2]

here, a= 93, b = 107

(100 - 7)3 - (100 + 7 )3 = -2[73 + 3 (100 )2(7)]

= -2(343 + 210000]

= -2(210343]

= -420686

The va lue o f 933 - 1073 = -420686

Q15. I f x + ^ = 3, calculate x 2 + + \ , x i + \X x a X4

S o l:

Given, x + — = 3X

We know th a t (x + y )2 = x 2 + y2 + 2xy

( x + ^ ) 2 = x 2 + ^ + ( 2 * * * 1 )

32 = x 2 + + 2X2

9 - 2 = x 2 + 4X2

x2+ 4 = 7

squaring on bo th s ides

( x2 + ^ ) 2 = 72

x4 + A . + 2 * x2 * ^ = 4 9a;4 x 1

X4 + J j . + 2 = 49x i

x4 + A - = 4 9 - 2X 4

x4 + 4 - = 4 7

a g a in , cub ing on bo th s ides

(x + ^ ) 3 = 33

x 3 + + 3 x *—(x + —) = 27

x 3 + ± + (3 *3 ) = 27

x 3 + J _ + g = 2 7x »

x 3 + \ = 2 7 - 9X A

x3+ \ = 1 8X s

hence, th e va lues are x 2 + = 7, x4 + -V = 47, x 3 + = 1 8X1 x 4 X S

Q16. If x4 + —7 = 194, calculate x 2 + ^ , x 3 + \ , x + ~X 4 X £ X 3 x

S o l:

Given,

x4 + 4 - = 194 --------1X 4

add and su b tra c t (2 *x 2 * - ^ ) on le ft s ide in above given equationX 1

x4 + —7- + (2 * x 2 * ^ ) - 2 (2 *x 2 * ± ) = 1 9 4

x4 + J_ + (2*a:2 * ^ ) - 2 = 194

(x2 + ~ j )2 “ 2 =194

(x2 + ^ ) 2 = 1 9 4 + 2

(x2+ ^ ) 2 = 196

(x2 + ^ ) = V ^ 9 6

(x2+ Jj ) = 14 -------- 2

A dd and su b tra c t (2 *x* on le ft s ide in eq 2

(x2 + ^ ) + ( 2 * x * i ) - ( 2 * x * | ) = 14

( x + ± ) 2 - 2 = 14

( x + i )2 = 1 4 + 2

( x + ^ ) 2 = 1 6

( x + i ) = 7 1 6

(x+ 7 ) =4 ------- 3

Now, cub ing eq 3 on bo th s ides

( x + i )3 = 4 3


x3+ ^ t + 3*x* ^ ( x + ^ ) = 64

x 3 + p r + (3 *4 ) = 64

x 3 + J = = 6 4 - 1 2X 3

x 3 + = 52X A

hence, th e va lues o f (x2 + ■ ■)2 = 196, (x + - j ) = 4 , x 3 + ^ = 5 2

Q17. Find the values o f 27x3 + By3 , if

(a) 3x + 2y = 14 and xy = 8

(b) 3x + 2y = 20 and xy = ^

Sol:

(a) Given, 3x + 2y = 14 and xy = 8

cub ing on both s ides

(3x + 2y)3 = 14 3


27x3 + 8y3 + 3 (3x)(2y)(3x + 2y) = 2744

27x3 + 8y3 + 18xy(3x + 2y) = 2744

27x3 + 8y3 + 18(8)(14) = 2744

27x3 + 8y3 + 2016 = 2744

27x3 + 8y3 = 2 7 4 4 - 2 0 1 6

27x3 + 8y3 = 728

Hence, the value o f 27x3 + Sy3 = 728

(b) Given, 3x + 2y = 20 and xy = - y

cub ing on both s ides

(3x + 2y)3 = 203


27x3 + 8y3 + 3 (3x)(2y)(3x + 2y) = 8000

27x3 + 8y3 + 18xy(3x + 2y) = 8000

27x3 + 8y3 + 18 ( ^ ) ( 2 0 ) = 8000

27x3 + 8y3 + 560 = 8000

27x3 + 8y3 = 8000 - 560

27x3 + 8y3 = 7440

Hence, the value o f 27x3 + 8y3 = 7440

Q18. Find the value o f 64X3 - 1 25z3 , if 4x - 5z = 16 and xz = 12

Sol:

Given, 64x3 - 125z3

Here, 4 x - 5z = 16 and xz = 12

Cubing 4x - 5z = 16 on bo th s ides

(4x - 5z)3 = 163

We know tha t, (a - b )3 = a3 - b3 - 3ab(a - b)

(4x)3 - (5z)3 - 3 (4x)(5z)(4x - 5z) = 163

64x3 - 125z3 - 60 (xz)(16) = 4096

64x3 - 125z3 - 60(12 )(1 6) = 4096

64x3 - 125z3 - 11520 = 4096

64x3 - 125z3 = 4 096 + 11520

64x3 - 125z3 = 1 5 6 1 6

The va lue o f 64x3 - 125z3 = 15616

Q19. If x - ^ = 3 + 2 \ / 2 , Find the value of x3 - ^

S o l:

Given, x - ^ = 3 + 2 \ / 2

Cubing x - = 3 + 2y/2 on both s ides


(x - I ) 3 = (3 + 2 ^ ) 3

x 3 - ± - 3 * x *± (x - | ) = 3 2 + (2-y/2)3+ 3 *3 *2 V ^ ( 3 + 2 i / 2 )

x 3 - 4 - " 3 (3 + 2 v ^ ) = 27 + 16 \ / 2 + 18 v ^ ( 3 + 2-s/2)

x 3 - 4 r = 27 + 16-v/2 + 5 4 + 72 + 9 + 6 -\/2X A

x 3 - ^ = 108 + 76 %/2X s

hence, th e va lue o f x3 - = 108 + 7 6 1/2X *


RD Sharma Solutions Class 9 Chapter 4 Ex 4.4Q1. Find the following products

(a) (3x + 2y)(9x2 - 6xy + 4y2)

(b) (4x - 5y)(16x2 + 20xy + 25y2)

(c) (7p4 + q)(49p8 - 7p4q + q2)

(d ) ( f +2y)(

T - x y + 4y2)

( 3 5 \ / 9 . 25 , 15 \

(f)(3 + | ) (9 -^ + f )

(g) ( ! + 3 a : ) ( £ + 9 a ! 2-6 )

(h) ( § - 2 * 2)(-^- + 4 x 4- 6 x )

(i) (1 - x)(1 + x + xA{2})

Q)(1 +x)(1 - x + xA{2})

(^(a^-lKa^ + ^ + l)

(I) ( x 2 + l ) ( x 6- x 3 + 1)

Sol:

(a) (3x + 2y)(9x2 - 6xy + 4y2)

Given, (3x + 2y)(9x2 - 6xy + 4y2)

We know that, a3 + b3 = (a + b)(a2 + b2 - ab)

(3x + 2y)(9x2 - 6xy + 4y2) can we written as

=> (3x + 2y)[(3x)2 - (3x)(2y) + (2y)2)]

=> (3x)3 + (2y)3

=> 27x3 + 8y3

Hence, the value o f (3x + 2y)(9x2 - 6xy + 4y2) = 27x3 + 8y3

(b) (4x - 5y)(16x2 + 20xy + 25y2)

Given, (4x - 5y)(16x2 + 20xy + 25y2)

We know that, a3 - b3 = (a - b)(a2 + b2 + ab)

(4x - 5y)(16x2 + 20xy + 25y2) can we written as

=> (4x - 5y)[(4x)2 + (4x)(5y) + (5y)2)]

=> (4x)3 - (5y)3

=> 16x3 - 25y3

Hence, the value o f (4x - 5y)(16x2 + 20xy + 25y2) = 16x3 - 25y3

(c) (7p4 + q)(49p8 - 7p4q + q2)

Given, (7p4 + q)(49p8 - 7p4q + q2)


(7p4 + q)(49p8 - 7p4q + q2) can be written as

=> (7p4 + q)[(7p4)2 - (7p4)(q) + (q)2)]

=> (7p4)3 + (q)3=> 343p12 + q3

Hence, the value o f (7p4 + q)(49p8 - 7p4q + q2) = 343p12 + q3

(d) ( f + 2y)( - xy + 4y2)

Sol:

Given, ( f + 2 y ) ( ^ - x y + 4y2)


( f + 2y)( ^ - xy + 4y2) can be written as

= > ( f + 2 3/) [ ( f ) 2- f ( 2 t /) + (2t/)2]

= > ( f ) 3 + (2 j, )3

=> x + 8y 3

* ) ( * + 7 + S *)

Sol:

G iv e n , ( | - f ) ( A + l | + i | )


/_3__5_\/_9__ |_ 25_ . 15 \' x y )\ x2 ' y 2 ' x y )

Can be written as,

= > ( ! - f ) ( ! ) 2+(f)2+ ( ! ) ( f )

=>(f)3- ( f )3

= > ( § ) - ( ? )

Hence, the value o f ( * - £ ) ( £ + f + § ) = ( f M f )

( f) (3 + | ) ( 9 - f + f )

Sol:

Given,(3 + f ) ( 9 - f + f )


(3 + § -)(9 - + J r ) can be written as,

= > ( 3 + § ) [ ( 3 2) - 3 ( f ) + ( § ) 2]

=>(3)3 + ( | ) 3= > 2 7+ * f

Hence, the value o f (3 + | ) ( 9 - £ + f ) is 27 + ^

( f l ) ( ! + 3 » ) ( ^ + 9 * 2- 6 )

Sol:

Given, ( J + 3a;)( J r + 9a;2- 6 )


(■J + 3a:)(-J- + 9a:2- 6) can be written as,

= > ( ! + 3 a : ) [ ( f ) 2 + (3a:)2- ( | ) ( 3 a : ) ]

= > ( | ) 3 + (3 x )3

= > J r + 9 * 3

Hence, the value o f (-J + 3a;) ( - ^ + 9a:2- 6) is J r + 9a:3

(h ) ( f - 2 a : 2) ( ^ - + 4 a : 4-6 a ;)

Sol:

Given, ( J -2 a ;2) ( J r + 4a:4-6 a :)


(■ § -2a:2)(-J- + 4a:4-6 a :) can be written as,

= > ( ! - 2 * 2) [ ( ! ) 2 + (2a;2)2- ( | ) ( 2 a : 2)]

= > ( f - 2 a ;2) [ ( J r ) + 4 a :4- ( f ) ( 2 a ; 2)]

= > ( f ) 3- (2 a :2) 3

=> ? j - 8 x 6

Hence, ( J - 2 a ;2)(-J- + 4a:4-6 a :) is | j - 8 a : 6

(i) (1 - x)(1 + x + xA{2})

Sol:

Given, (1 - x)(1 + x + xA{2})


(1 - x)(1 + x + xA{2}) can be written as,

=>(1 -x ) [(1 2 + (1)(x)+x2)]

=>(1)3 - ( x ) 3

=> 1 - X3

Hence, the value o f (1 - x)(1 + x + xA{2}) is 1 - x3

( j) (1 +x)(1 - x + xA{2})

Sol:

Given, (1 + x)(1 - x + xA{2})


(1 + x)(1 - x + xA{2}) can be written as,

=>(1+x)[(12 -(1 )(x ) + x2)]

=> (1)3 + (x)3

=> 1 + X3

Hence, the value o f (1 + x)(1 + x - xA{2}) is 1 + x3

(k) (x 2- l) (a ;4 + x 2 + 1)

Sol:

Given, (x 2- 1) (a:4 + x 2 + 1)


(a:2- l) (a :4 + as2 + 1) can be written as,

=>(x2-1 ) [(x 2)2 - 1 2 + (x2)(1)]

=> (x2)3 - 13

=> x6 - 1

Hence, (a;2- l) (a ;4 + a;2 + 1) is x6 - 1

(l) ( x2 + l ) ( x 6- x 3 + 1)

Sol:

Given, ( x2 + l ) ( x 6- x 3 + 1)


(a:2 + l) (a :6-a ;3 + 1) can be written as,

=>(x3 + 1)[(x3)2 - ( x 3)(1) + 12]

=> (x3)3 + 13

=> X9 + 1

Hence, the value o f (a;2 + l) (a ;6-a :3 + 1) is x9 +1

Q2. Find x= 3 and y = -1, Find the values o f each of the following using in identity:

(a) (9x2 - 4x2)(81 y4 + 36x2y2 + 16x4)

/la / 3 5 \ / 9 , 2 5 , 1 5 \

r~\rx v \ ( ^ i v2 xv\(c ) ( 7 + 3 ) ( 49 + 9 21 )

/ / x y x/ a:2 , v2 , xv(d) ( l “ 3 ) ( l6 + T + 2 l)

(e) (| + 5 x ) ( f - 2 5 + 25a:2)

Sol:

(a) (9x2 - 4x2)(81 y4 + 36x2y2 + 16x4)


(9x2 - 4x2)(81 y4 + 36x2y2 + 16X4) can be written as,

=> (9x2 - 4x2)[(9y2)2 + (9)(4)x2y2 + (4x2)2]

=> (9y2)3 - (4x2)3

=> 729y6 - 64x6

Substitute the value x = 3, y = -1 in 729y6 - 64x6 we get,

=> 729y6 - 64x6

=> 729(-1)6 - 64(3)6

=> 729(1) - 64(729)

=> 729 - 46656

=> -45927

Hence, the product value o f (9x2 - 4x2)(81y4 + 36x2y2 + 16x4) = -45927

■ 2t) | 15 \«2 “T xv )xy J

Sol:

Given,


/_3_ 5\/_9__ |_ 25 _|_ 15 \\x y ) y x2 ^ ~ y2 ^ ~ Xy)

Can be written as,

= > ( ! - f ) [ ( f ) 2 + ( f ) 2 + ( ! ) ( ! ) ]

= > ( | ) 3- ( f ) 3

= > ( f ) - ( # ) - - • >

Substitute x = 3 in eq 1

=> 0

Hence, the value ° f ( f - f ) ( J r + | r + i f ) is 0

/„W I , V \( X* , V* XV\(cH t + j ) ( i 8 + y _ n )

Given,


(2L+ V ) ( J + £ - 2 . )

Can be written as,

= > ( f + ! ) [ ( f ) 2 + ( | ) 2- ( f ) ( | ) ]

= > ( f ) 3 + ( | ) 3

=>(^) + ( S ) - - 1Substitute x = 3, y = -1 in eq 1

V 343 > ~ \ 27 >

= > ( -» W J_)V 343 > 27 7

Taking least common multiple, we get

27*27 1*343-> 343*27 27*343

-> 729 3439261 9261

__ 729-343 9261

386 - 9261

Hence, the value o f ( f + f ) ( f f +

/a\ <x v\rx 2 _i » 2 W ( 4 3 ) ( 16 9 21 )

xy_ \ _ 386 21 > ~ 9261

Sol:

Given,


( E - l ) ( £ + £ + 2 )1 4 3 16 T 9 T 21 )

Can be written as,

= > ( f - ! ) [ ( f ) 2 + ( ! ) 2 + ( f ) ( 1 ) ]

= > ( f ) 3- ( ! ) 3

V 64 / ' 27 >

Substitute x = 3, y = -1 in eq 1

V 343 ' v 27 '

= > (I ) + (w)Taking least common multiple, we get

_ > 27*27 , 1*6464*27 ~l~ 27*64

_ 729 , 64" 1728 1728

729+64 -> 1728

_ 7939261

Hence, the value o f + £ + £ ) = ^

(e)(| + 5 x ) ( f - 2 5 + 25a:2)

Sol:


(■| + 5 x )(-p — 25 + 25a;2) can be written as,

= > ( ! + 5x) [ ( ! )2 + (5x )2- ( ! ) ( 5a;)]

= > ( | ) 3 + (5 x )3

=> + 125a:3 ----- 1

Substitute x = 3, in eq 1

=> ™ + 125(3)3

=>2§ + 125 * 27

=>2§ + 3375

Taking least common multiple, we get

__ 125 . 3375*27 27 27*1

_ 125 , 9112527 27

125+91125 -> 27

_ 91250 25

Hence, the value o f ( - | + 5a;) (-p— 25 + 25a;2) is 91230

Q3. If a + b = 10 and ab = 16, find the value o f a2 - ab + b2 and a2 + ab + b2

Sol:

Given, a + b = 10, ab = 16

We know that, (a + b)3 = a3 + b3 + 3ab(a + b)

=> a3 + b3 = (a + b)3 - 3ab(a + b)

=> a3 + b3 = (10)3 - 3(16)(10)

=> a3 + b3 = 1000 - 480

=> a3 + b3 = 520

Substitute, a3 + b3 = 520, a + b = 10 in a3 + b3 = (a + b)(a2 + b2 - ab)

a3 + b3 = (a + b)(a2 + b2 - ab)

520 = 10(a2 + b2 - ab)

^ = ( a 2 + b2 -a b )

=> (a2 + b2 - ab) = 52

Now, we need to find (a2 + b2 + ab)

Add and subtract 2ab in a2 + b2 + ab

=> a2 + b2 + ab - 2ab + 2ab

=> (a + b)2 - ab

Substitute a + b = 10, ab

=> a2 + b2 + ab = 102 - 16

= 1 0 0 -1 6

= 84

Hence, the values of (a2 + b2 - ab) = 52 and (a2 + b2 + ab) = 84

Q4. If a + b = 8 and ab = 6, find the value o f a3 + b3

Sol:

Given, a + b = 8 and ab = 6

We know that, a3 + b3 = (a + b)3 - 3ab(a + b)

=> a3 + b3 = (a + b)3 - 3ab(a + b)

=> a3 + b3 = (8)3 - 3(6)(8)

=> a3 + b3 = 512 - 144

=> a3 + b3 = 368

Hence, the value o f a3 + b3 is 368

Q5. If a - b = 6 and ab = 20, find the value of a3 - b3

Sol:

Given, a - b = 6 and ab = 20

We know that, a3 - b3 = (a - b)3 + 3ab(a - b)

=> a3 - b3 = (a - b)3 + 3ab(a - b)

=> a3 - b3 = (6)3 + 3(20)(6)

=> a3 - b3 = 216 + 360

=> a3 - b3 = 576

Hence, the value o f a3 - b3 is 576

Q6. If x = -2 and y = 1, by using an identity find the value o f the following:

(a) (4y2 - 9x2)(1 6y4 + 36x2y2 + 81 x4)

(b) ( | - f ) ( ^ + T + 1)(c) (by + y )(25 j/2- 7 5 + ^ j r )

Sol:

Given,

(a) (4y2 - 9x2)(1 6y* + 36x2y2 + 81 x4)


(4y2 - 9x2)(1 by4 + 36x2y2 + 81 x4) can be written as,

=> (4y2 - 9x2)[(4x)2 + 4y2*9x2 + (9x2)2)

=> (4y2)3 - (9x2)3

=> 64y6 - 729x6 --------1

Substitute x = -2 and y = 1 in eq 1

=> 64y6 - 729x6

=> 64(1 )6 - 729(-2)6

=> 64 - 729(64)

=> 64(1 - 729)

=> 64(-728)

=> -46592

Hence, the value o f (4y2 - 9x2)(1 by4 + 36x2y2 + 81 x4) is -46592

(b) ( f - f ) ( ^ + T + 1)

here x = -2


( ! - f-)(Jr + T + 1) can be witten as-

= > ( M ) [ ( f ) 2 + ( f ) 2 + ( ! ) ( ! ) ]

= > ( | ) 3- ( f ) 3

= > ( £ ) - ( £ ) — 1

Substitute x = -2 in eq 1

=> -1 +1

=> 0

Hence, the value o f ( | — f )(-p- + y + 1) is 0

( c ) ( 5 y + f ) ( 2 5 y 2- 7 5 + ^ )

Sol:


(5 y + y ) ( 2 5 t / 2- 7 5 + ^y -) can be written as,y y

= > (5 y + f) [(5 y )2 + ( f ) 2 - ( 5 y ) ( f ) ]

=>(5y)3 + ( f ) 3=>125y3 + ( ^ ) ------ 1

y3

Substitute y = 1 ineq 1

=>125(1)3 + ( ^ )

=>125 + 3375

=> 3500

Hence, the value o f (5y + y ) (25s/2— 75 + ) is 3500.

EXERCISE 4.5

Q1. Find the following products:

(a) (3x + 2y + 2z)(9x2 + 4y2 + 4z2 - 6xy - 4yz - 6zx)

(b) (4x - 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz - 8zx)

(c) (2a - 3b - 2c)(4a2 + 9b2 + 4c2 + 6ab - 6bc + 4ca)

(d) (3x - 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy - 15zx + 20yz)

Sol:

Given,

(a) (3x + 2y + 2z)(9x2 + 4y2 + 4z2 - 6xy - 4yz - 6zx)

we know that,

x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)

so,

(3x + 2y + 2z)(9x2 + 4y2 + 4z2 - 6xy - 4yz - 6zx) = (3x)3 + (2y)3 + (2z)3 - 3(3x)(2y)(2z)

= 27x3 + 8y3 + 8z3 - 36xyz

Hence, the value o f (3x + 2y + 2z)(9x2 + 4y2 + 4z2 - 6xy - 4yz - 6zx) is 27x3 + Sy3 + 8z3 - 36xyz

(b) (4x - 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz - 8zx)

we know that,


so.

(4x - 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz - 8zx) = (4x)3 + (-3y)3 + (2z)3 -3(4x)(-3y)(2z)

= 64x3 - 27y3+ 8z3 + 72xyz

Hence, the value o f (4x - 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz - 8zx) is 64x3 - 27y3+ 8z3 + 72xyz

(c) (2a - 3b - 2c)(4a2 + 9b2 + 4c2 + 6ab - 6bc + 4ca)

we know that,


so,

(2a - 3b - 2c)(4a2 + 9b2 + 4c2 + 6ab - 6bc + 4ca) = (2a)3 + (-3b)3 + (-2c)3 - 3(2a)(-3b)(-2c)

= 8a3 - 27b3 - 8c3 - 36abc

Hence, the value o f (c) (2a - 3b - 2c)(4a2 + 9b2 + 4c2 + 6ab - 6bc + 4ca) is 8a3 - 27b3 - 8c3 - 36abc

(d) (3x - 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy - 15zx + 20yz)

we know that,


so,

(3x - 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy - 15zx + 20yz) = (3x)3 + (-4y)3 + (5z)3 -3(3x)(-4y)(5z)

= 27x3 - 64y3 + 125z3 + 180xyz

Hence, the value o f (3x - 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy - 15zx + 20yz) is 27x3 - 64y3 + 125z3 + 180xyz

Q2. I fx + y + z = 8 and xy + yz + zx = 20, Find the value of x3 + y3 + z3 - 3xyz

Sol:given, x + y + z = 8 and xy + yz + zx = 20

We know that,

(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)

(x + y + z)2 = x2 + y2 + z2 + 2(20)

(x + y + z)2 = x2 + y2 + z2 + 40

82 = x2 + y2 + z2 + 40

64 - 40 = x2 + y2 + z2

x2 + y2 + z2 = 24

we know that,


x3 + y3 + z3 - 3xyz = (x + y + z)[(x2 + y2 + z2) - (xy + yz + zx)]

here, x + y + z = 8, xy + yz + zx = 20, x2 + y2 + z2 = 24

x3 + y3 + z3 - 3xyz = 8[(24 - 20)]

= 8 * 4

= 32

Hence, the value o f x3 + y3 + z3 - 3xyz is 32

Q3. Ifa + b + c = 9 and ab + be + ca = 26, Find the value o f a3 + b3 + c3 - 3abc

Sol:

Given, a + b + c = 9 and ab + be + ca = 26

We know that,

(a + b + c)2 = a2 + b2 + c2 + 2(ab + be + ca)

(a + b + c)2 = a2 + b2 + c2 + 2(26)

(a + b + c)2 = a2 + b2 + c2 + 52

92 = a2 + b2 + c2 + 52

81 - 52 = a2 + b2 + c2

a2 + b2 + c2 = 29

we know that.

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)

a3 + b3 + c3 - 3abc = (a + b + c)[(a2 + b2 + c2) - (ab + be + ca)]

here, a + b + c = 9, ab + be + ca = 26, a2 + b2 + c2 = 29

a3 + b3 + c3 - 3abc = 9[(29 - 26)]

= 9 * 3

= 27

Hence, the value o f a3 + b3 + c3 - 3abc is 27

Q4. Ifa + b + c = 9 and a2 + b2 + c2 = 35, Find the value o f a3 + b3 + c3 - 3abc

Sol:

Given, a + b + c = 9 and a2 + b2 + c2 = 35

We know that,

(a + b + c)2 = a2 + b2 + c2 + 2(ab + be + ca)

92 = 35 + 2(ab + be + ca)

81 = 35 + 2(ab + be + ca)

81 - 35 = 2(ab + be + ca)

■y = ab + be + ca

ab + be + ca = 23

we know that,


a3 + b3 + c3 - 3abc = (a + b + c)[(a2 + b2 + c2) - (ab + be + ca)]

here, a + b + c = 9, ab + be + ca = 23, a2 + b2 + c2 = 35

a3 + b3 + c3 - 3abc = 9[(35 - 23)]

= 9 *1 2

= 108

Hence, the value o f a3 + b3 + c3 - 3abc is 108

Q5. Evaluate:

(a) 253 - 753 + 503

(b) 483 - 303 - 183

(c ) ( ^ ) 3 + ( | ) 3- ( ! ) 3

(d) (0.2)3 - (0.3)3 + (0.1 )3

Sol:

Given,

(a) 253 - 753 + 503

we know that,


here, a = 25, b = -75, c = 50


a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 - ab - be - ca) + 3abc

a3 + b3 + c3 = (25 - 75 + 50)(a2 + b2 + c2 - ab - be - ca) + 3abc

a3 + b3 + c3 = (0)(a2 + b2 + c2 - ab - be - ca) + 3abc

a3 + b3 + c3 = 3abc

253 + (-75)3 + 503 = 3abc

= 3(25)(-75)(50)

= -281250

Hence, the value 253 + (-75)3 + 503 = -281250

(b) 483 - 303 - 183

we know that,


here, a = 48, b = -30, c = -18



a3 + b3 + c3 = (48 - 30 - 18)(a2 + b2 + c2 - ab - be - ca) + 3abc


a3 + b3 + c3 = 3abc

483 + (-30)3 + (-18)3 = 3abc

= 3(48)(-30)(-18)

= 77760

Hence, the value 483 + (-30)3 + (-18)3 = 77760

( c ) ( i ) 3 + ( | ) 3- ( ! ) 3

we know that,


here, a= | , b = | , c = ^



a3 + b3 + c3 = (- j +-| - - |)( a2 + b2 + c2 - ab - be - ca) + 3abc

by using least common multiple

a3 + b3 + c3 = ( ^ | + f j§ ) ( a2 + b2 + c2 - ab - be - ca) + 3abc

a3 + b3 + c3 = + j 2~ - j f ) ( a2 + b2 + c2 - ab - be - ca) + 3abc

a3 + b3 + c3 = 0( a2 + b2 + c2 - ab - be - ca) + 3abc

a3 + b3 + c3 = 3abc

( | ) 3 + ( l ) 3 - ( ^ ) 3 = 3 * f 4 *- 56

Hence, the value o f ( | )3 + ( f ) 3 - ( | ) 3 is

(d) (0.2)3 - (0.3)3 + (0.1 )3

we know that,


here, a = 0.2, b = 0.3, c = 0.1



a3 + b3 + c3 = (0.2 - 0.3 + 0.1 )(a2 + b2 + c2 - ab - be - ca) + 3abc


a3 + b3 + c3 = 3abc

(0.2)3 - (0.3)3 + (0.1 )3 = 3abc

= 3(0.2)(-0.3)(0.1)

= -0.018

Hence, the value (0.2)3 - (0.3)3 + (0.1 )3 is 0.018

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4x2 + — b2 — 2ab a = 2x, b€¦ · RD Sharma Solutions Class 9 Algebraic Identities Ex 4.1 RD Sharma Solutions Class 9 Chapter 4 Ex 4.1 1) Evaluate each of the following using