Chapter 03.02 Solution of Cubic Equations After reading this chapter, you should be able to:

1. find the exact solution of a general cubic equation. How to Find the Exact Solution of a General Cubic Equation In this chapter, we are going to find the exact solution of a general cubic equation

023 =+++ dcxbxax (1) To find the roots of Equation (1), we first get rid of the quadratic term ( )2x by making the substitution

abyx3

−= (2)

to obtain

0333

23

=+⎟⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ − d

abyc

abyb

abya (3)

Expanding Equation (3) and simplifying, we obtain the following equation

0327

23 2

323 =⎟⎟

⎠

⎞⎜⎜⎝

⎛−++⎟⎟

⎠

⎞⎜⎜⎝

⎛−+

abc

abdy

abcay (4)

Equation (4) is called the depressed cubic since the quadratic term is absent. Having the equation in this form makes it easier to solve for the roots of the cubic equation (Click here to know the history behind solving cubic equations exactly). First, convert the depressed cubic Equation (4) into the form

0327

213

12

323 =⎟⎟

⎠

⎞⎜⎜⎝

⎛−++⎟⎟

⎠

⎞⎜⎜⎝

⎛−+

abc

abd

ay

abc

ay

03 =++ feyy (5) where

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

abc

ae

31 2

03.02.1

03.02.2 Chapter 03.02

⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

abc

abd

af

32721

2

3

Now, reduce the above equation using Vieta’s substitution

zszy += (6)

For the time being, the constant is undefined. Substituting into the depressed cubic Equation (5), we get

s

03

=+⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ + f

zsze

zsz (7)

Expanding out and multiplying both sides by , we get 3z( ) ( ) 033 32346 =++++++ szessfzzesz (8)

Now, let 3es −= ( is no longer undefined) to simplify the equation into a tri-quadratic

equation.

s

027

336 =−+

efzz (9)

By making one more substitution, , we now have a general quadratic equation which can be solved using the quadratic formula.

3zw =

027

32 =−+

efww (10)

Once you obtain the solution to this quadratic equation, back substitute using the previous substitutions to obtain the roots to the general cubic equation.

xyzw →→→ where we assumed

3zw = (11)

zszy +=

3es −= (12)

abyx3

−=

Note: You will get two roots for as Equation (10) is a quadratic equation. Using

would then give you three roots for each of the two roots of , hence giving you six root values for

w(11)Equation w

z . But the six root values of z would give you six values of ( ); but three values of will be identical to the other three. So one gets only three values of , and hence three values of

y(6)Equation y

y x . (Equation (2)) Example 1 Find the roots of the following cubic equation.

080369 23 =−+− xxx

Solution of Cubic Equations 03.02.3

Solution

For the general form given by Equation (1) 023 =+++ dcxbxax

we have 1=a , , , 9−=b 36=c 80−=d

in 080369 23 =−+− xxx (E1-1)

Equation (E1-1) is reduced to 03 =++ feyy

where

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

abc

ae

31 2

( )( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−=

13936

11 2

9=and

⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

abc

abd

af

32721

2

3

( )( )

( )( )( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

−+−=

13369

1279280

11

2

3

26−=giving

02693 =−+ yy (E1-2) For the general form given by Equation (5)

03 =++ feyy we have

9=e , 26−=fin Equation (E1-2). From Equation (12)

3es −=

39

−=

3−=From Equation (10)

027

32 =−+

efww

027926

32 =−− ww

027262 =−− ww

03.02.4 Chapter 03.02

where 3zw =

and

zszy +=

z

z 3−=

( ) ( ) ( )( )( )12

27142626 2 −−−±−−=w

1,27 −=The solution is

271 =w 12 −=w

Since 3zw = wz =3

For 1ww =

13 wz =

27= 027 ie=Since

3zw = ( ) ααθ iii euuere 333

== ( ) ( ααθθ 3sin3cossincos 3 iuir +=+ )

resulting in 3ur =

αθ 3coscos = αθ 3sinsin =

Since θsin and θcos are periodic of π2 , kπθα 23 +=

32 kπθα +

=

k will take the value of 0, 1 and 2 before repeating the same values of α . So,

2 ,1 ,0 ,32

=+

= kkπθα

31θα =

( )32

2πθα +

=

Solution of Cubic Equations 03.02.5

( )34

3πθα +

=

So roots of are 3zw =

⎟⎠⎞

⎜⎝⎛ +=

3sin

3cos3

1

1θθ irz

⎟⎠⎞

⎜⎝⎛ +

++

=32sin

32cos3

1

2πθπθ irz

⎟⎠⎞

⎜⎝⎛ +

++

=34sin

34cos3

1

3πθπθ irz

gives

( ) ⎟⎠⎞

⎜⎝⎛ +=

30sin

30cos27 3/1

1 iz

3=

( ) ⎟⎠⎞

⎜⎝⎛ +

++

=320sin

320cos27 3/1

2ππ iz

⎟⎠⎞

⎜⎝⎛ +=

32sin

32cos3 ππ i

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

23

213 i

2

3323 i+−=

( ) ⎟⎠⎞

⎜⎝⎛ +

++

=340sin

340cos27 3/1

3ππ iz

⎟⎠⎞

⎜⎝⎛ +=

34sin

34cos3 ππ i

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

23

213 i

2

3323 i−−=

Since

zzy 3−=

111

3z

zy −=

333−=

2=

03.02.6 Chapter 03.02

222

3z

zy −=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

233

23

32

3323

ii

31335

ii

+−+

−=

3131

31335

ii

ii

−−−−

×+−+

−=

321 i+−=

333

3z

zy −=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

233

23

32

3323

ii

31335

ii

+−

=

3131

31335

ii

ii

−−

×+−

=

321 i−−= Since

3+= yx 311 += yx

32 += 5=

322 += yx ( ) 3321 ++−= i 322 i+=

333 += yx

( ) 3321 +−−= i 322 i−=

The roots of the original cubic equation

080369 23 =−+− xxx are and , that is, , , 21 xx 3x

5 , 322 i+ , 322 i− Verifying

Solution of Cubic Equations 03.02.7

( ) ( )( ) ( )( ) 03223225 =−−+−− ixixx gives

080369 23 =−+− xxx Using

12 −=w would yield the same values of the three roots of the equation. Try it. Example 2 Find the roots of the following cubic equation 0104.203.0 623 =×+− −xxSolution For the general form

023 =+++ dcxbxax 6104.2 ,0 ,03.0 ,1 −×==−== dcba

Depress the cubic equation by letting (Equation (2))

abyx

3−=

( )( )1303.0 −

−= y

01.0 += y Substituting the above equation into the cubic equation and simplifying, we get

( ) ( ) 0104103 743 =×+×− −− yy That gives and for Equation (5), that is, . 4103 −×−=e 7104 −×=f 03 =++ feyyNow, solve the depressed cubic equation by using Vieta’s substitution as

zszy +=

to obtain ( ) ( ) ( ) 010331041033 32437446 =+×−+×+×−+ −−− szsszzsz

Letting 4

4

103103

3−

−

=×−

−=−=es

we get the following tri-quadratic equation ( ) 0101104 12376 =×+×+ −− zz

Using the following conversion, , we get a general quadratic equation 3zw =( ) ( ) 0101104 1272 =×+×+ −− ww

Using the quadratic equation, the solutions for are w

( ) ( )( )( )12

10114104104 12277 −−− ×−×±×−=w

giving

03.02.8 Chapter 03.02

( )771 1037979589711.9102 −− ×+×−= iw

( )772 1037979589711.9102 −− ×−×−= iw

Each solution of yields three values of 3zw = z . The three values of z from are in rectangular form.

1w

Since 3zw =

Then

31

wz = Let

( ) θθθ ireirw =+= sincos then

( ) ααα iueiuz =+= sincos This gives

3zw = ( ) ααθ iii euuere 333

== ( ) ( ααθθ 3sin3cossincos 3 iuir +=+ )

resulting in 3ur = αθ 3coscos = αθ 3sinsin = Since θsin and θcos are periodic of π2 ,

kπθα 23 +=

32 kπθα +

=

k will take the value of 0, 1 and 2 before repeating the same values of α . So,

2 ,1 ,0 ,32

=+

= kkπθα

31θα =

( )32

2πθα +

=

( )34

3πθα +

=

So the roots of are 3zw =

⎟⎠⎞

⎜⎝⎛ +=

3sin

3cos3

1

1θθ irz

⎟⎠⎞

⎜⎝⎛ +

++

=32sin

32cos3

1

2πθπθ irz

Solution of Cubic Equations 03.02.9

⎟⎠⎞

⎜⎝⎛ +

++

=34sin

34cos3

1

3πθπθ irz

So for ( )77

1 1037979589711.9102 −− ×+×−= iw

( ) ( )2727 1037979589711.9102 −− ×+×−=r 6101 −×=

7

71

1021037979589711.9tan −

−−

×−×

=θ

(2nd quadrant because (the numerator) is positive and 772154248.1= y x (the denominator) is negative)

( ) ⎟⎠⎞

⎜⎝⎛ +×= −

3772154248.1sin

3772154248.1cos101 3

16

1 iz

350055695756.0170083054095.0 i+=

( ) ⎟⎠⎞

⎜⎝⎛ +

++

×= −

32772154248.1sin

32772154248.1cos101 3

16

2ππ iz

150044079078.0460089760987.0 i+−=

( ) ⎟⎠⎞

⎜⎝⎛ +

++

×= −

34772154248.1sin

34772154248.1cos101 3

16

3ππ iz

480099774834.03130006706892.0 i−=Compiling

340055695756.0180083054095.01 iz += 140044079078.0460089760987.02 iz +−=

480099774834.01057068922852.6 43 iz −×= −

Similarly, the three values of z from in rectangular form are 2w340055695756.0180083054095.04 iz −=

140044079078.0460089760987.05 iz −−= 480099774834.01057068922852.6 4

6 iz +×= − Using Vieta’s substitution (Equation (6)),

zszy +=

( )z

zy4101 −×

+=

we back substitute to find three values for . yFor example, choosing

340055695756.0180083054095.01 iz += gives

340055695756.0180083054095.0101340055695756.0180083054095.0

4

1 iiy

+×

++=−

03.02.10 Chapter 03.02

40055695763.0180083054095.040055695763.0180083054095.0

40055695763.0780083054090.0101

340055695756.0180083054095.04

ii

i

i

−−

×+

×+

+=−

( )40055695763.0180083054095.0101101

340055695756.0180083054095.0

4

4

i

i

−××

+

+=

−

−

360166108190.0= The values of , and give 1z 2z 3z

360166108190.01 =y 90179521974.02 −=y

570013413784.03 =y respectively. The three other z values of , and give the same values as , and

, respectively. 4z 5z 6z 1y 2y

3yNow, using the substitution of

01.0+= yx the three roots of the given cubic equation are

01.0360166108190.01 +=x 360266108190.0=

01.090179521974.02 +−=x 90079521974.0−=

01.0570013413784.03 +=x 570113413784.0=

NONLINEAR EQUATIONS Topic Exact Solution to Cubic Equations Summary Textbook notes on finding the exact solution to a cubic

equation. Major General Engineering Authors Autar Kaw Last Revised July 3, 2009 Web Site http://numericalmethods.eng.usf.edu