Solve for x in complex numbers set: x^6 - 7x^3 - 8 = 0

Let x^3 be t:

We'll re-write
the equation:

t^2 - 7t - 8 =
0

We'll apply quadratic
formula:

t1 = [7+sqrt(49 +
32)]/2

t1 = (7+9)/2

t1 =
8

t2 = -1

x^3 = 8 =>
x^3 - 8 = 0

The difference of cubes returns the
product:

x^3 - 8 =(x-2)(x^2 + 2x +
4)

If x^3 - 8 = 0 => (x-2)(x^2 + 2x + 4) =
0

x - 2 = 0 => x =
2

x^2 + 2x + 4 = 0

We'll apply
quadratic formula:

x1 =
[-2+sqrt(4-16)]/2

x1 = (-2 +
2i*sqrt3)/2

x1 = -1+i*sqrt3

x2
= -1-i*sqrt3

x^3 = -1 => x^3 + 1 =
0

The difference of cubes returns the
product:

x^3 + 1 = (x+1)(x^2 - x +
1)

We'll cancel each factor:

x
+ 1 = 0 => x = -1

x^2 - x + 1 =
0

We'll apply quadratic
formula:

x1 =
[1+sqrt(1-4)]/2

x1 =
(1+isqrt3)/2

x2 =
(1-isqrt3)/2

The complex roots of equation
are: {-1 ; 2 ; -1+i*sqrt3 ; -1-i*sqrt3 ; (1+isqrt3)/2 ;
(1-isqrt3)/2}.