One for all, and all for one!

$x^2+x+1=0 \implies x^3+x^2+x=0 \\ \text{Subtract the Two equations} \\ \boxed{x^3=1}$

I used a simple factoring method:

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

$x^3-1=0$

$x^3 = \boxed{1}$

Clearly from the cube roots of unity, we have the two complex roots $\text {x}$ $=$ $\omega$ or $\omega^{2}$ ,

Now $x^{3}$ $=$ $\omega^{3}$ $=$ $1$ , by the product of the cube roots of unity.