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.**

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