Question 4625
I had started an answer on this problem but was called away from my computer before I could finish and post it.

Here's another approach to solving this.  First, recognise that, since this is 6th order polynomial, you can expect to find 6 roots.

1)  Factor the equation:  (x^3 + 8)(x^3 - 1) = 0
   Apply the zero products principle:  (x^3 + 8) = 0  or  (x^3 - 1) = 0

2) Notice that you now have the product of the sum of two cubes (x^3 + 2^3) and the difference of two cubes (x^3 - 1^3)

Factor the sum of two cubes:  x^3 + 8 = (x + 2)(x^2 - 2x + 4) = 0

So we have: (x + 2)(x^2 - 2x + 4) = 0;
x + 2 = 0; x = -2  or  
x^2 - 2x + 4 = 0; This doesn't factor so use the quadratic formula:

{{{x=(2+-sqrt(4-16))/(2)}}}
x = 1 + {{{sqrt(3)}}}i or x = 1 - {{{sqrt(3)}}}i

Factor the difference of two cubes:  x^3 - 1 = (x - 1)(x^2 + x + 1) = 0

So we have: (x - 1)(x^2 + x + 1) = 0
x - 1 = 0; x = 1 or
x^2 + x + 1 = 0  This doesn't factor so use the quadratic formula:

{{{x=(-1+-sqrt(1-4))/(2)}}}
x = -0.5 + 0.5{{{sqrt(3)}}}i or x = -0.5 - 0.5{{{sqrt(3)}}}i

The 6 roots are:
x = 1
x = -2
x = 1 + {{{sqrt(3)}}}i
x = 1 - {{{sqrt(3)}}}i
x = -0.5 + 0.5{{{sqrt(3)}}}i
x = -0.5 - 0.5{{{sqrt(3)}}}i