Find the roots of the equation x^6 - 9x^3 + 8 = 0

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We have to find the roots of x^6 - 9x^3 + 8 = 0.

x^6 - 9x^3 + 8 = 0

=> x^6 - 8x^3 – x^3 + 8 =0

=> x^3(x^3 - 8) - 1*(x^3 - 8) =0

=> (x^3 - 1) (x^3 - 8) =0

Now use the difference of cubes relation x^3 – y^3 = (x – y) (x^2 + xy +y^2)

=> (x -1) (x^2 + x +1) (x – 2) (x^2 + 2x + 4) = 0

For x – 1 = 0, we have x = 1

for x – 2 = 0, we have x = 2

To find the roots of x^2 + x +1 =0

we use the expressions for the roots of a quadratic equation ax^2 + bx +c = 0 which is [–b + sqrt (b^2 – 4ac)]/ 2a and [–b - sqrt (b^2 – 4ac)]/ 2a

Here the roots are [-1 – sqrt (1 – 4)]/ 2 = [-1 – sqrt (-3)]/2 = -1/2 – i*(sqrt 3)/2 and [-1 + sqrt (1 – 4)]/ 2 = [-1 + sqrt (-3)]/2 = -1/2 + i*(sqrt 3)/2

Similarly the roots of (x^2 + 2x + 4) are -1 + i*sqrt 3 and -1 - i*sqrt 3

**Therefore the roots are 1, 2, -1+i*sqrt 3 and -1-i*sqrt 3, -1/2 + i*(sqrt 3)/2 and -1/2 – i*(sqrt 3)/2.**

To find the roots of the equation x^6 - 9x^3 + 8 = 0

This is a quadratic equation in x^3.

So we can write the given equation as below:

t^2 -9t +8 = 0, where t = x^3.

t^2 -8t - t +8 = 0.

t(t-8) -1(1-8) = 0.

(t-8)(t-1) = 0.

Therefore t-8 = 0 , or t -1 = 0.

t-8 = 0 gives t=8, or x^3 = 8, so x = 2 the real root.

t-1 = 0 gives t = 1. So x^3 = 1. x= 1 , the real roots.

Therefore the real roots of the equation x^6 -9x^3+8 = 0 are x= 1 , x= 2.

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