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

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.