Is this Complex Enough for you?

$x$ is a cube root of unity:

$x^2+x+1 = \dfrac{x^3-1}{x-1} = 0 \implies x^3-1 = 0 \implies x^3 = 1$

So $x^6+\dfrac{1}{x^6} = (x^3)^2+\dfrac{1}{(x^3)^2} = 1^2+\dfrac{1}{1^2} = 1+1 = \boxed{2}$

Using the quadratic formula (with a=b=c=1) gives $x_{1}$ = - $\frac{1}{2}$ + $\frac{\sqrt{3}}{2}$ ${j}$ and $x_{2}$ = - $\frac{1}{2}$ - $\frac{\sqrt{3}}{2}$ ${j}$ , where ${j}$ = $\sqrt{-1}$ .

Using De Moivre's: $x^6_1$ = [- $\frac{1}{2}$ + $\frac{\sqrt{3}}{2}$ ${j}$ ]^6 = [Cos $\frac{2\pi}{3}$ + ${j}$ Sin $\frac{2\pi}{3}$ ]^6 = Cos4 $\pi$ + ${j}$ Sin 4 $\pi$ = 1.

Similarily: $x^6_2$ = [- $\frac{1}{2}$ - $\frac{\sqrt{3}}{2}$ ${j}$ ]^6 = [Cos $\frac{4\pi}{3}$ + ${j}$ Sin $\frac{4\pi}{3}$ ]^6 = Cos8 $\pi$ + ${j}$ Sin 8 $\pi$ = 1.

$\therefore$ $x^{6}$ + $\frac{1}{x^6}$ = 2

Since ${ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =-1$ .Cube it ${ { \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =-1 \right) } }^{ 3 }$ Produces ${ x }^{ 6 }+3\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) +\frac { 1 }{ { x }^{ 6 } } =-1$ Substitute ${ x }^{ 6 }+3\left( -1 \right) +\frac { 1 }{ { x }^{ 6 } } =-1$ ${ x }^{ 6 }+\frac { 1 }{ { x }^{ 6 } } =\boxed{2}$

Solution to the first equation is $(-1)^{2/3}$

Simply, putting the value in second equation, we get $\boxed{1+1=2}$

solution of equation 1 are omega and omega square put the value in the question asked so we get 1+1=2 answer