Why is $w^2$ a complex root for the equation $x^2+x+1=0$?

Question

"Multiplying both sides by $(x-1)$ gives $x^3-1=0$ which is the 3rd roots of unity. Its complex roots are also denoted as $w$ and $w^2$."

Answer 1

Its pretty obvious after you examime the geometric sum; $$x^2+x+1=\frac{x^3-1}{x-1}$$ In general $$\sum_{j=0}^{n-1} x^j=\frac{x^n-1}{x-1}$$

Answer 2

Let $w$ be a complex root of $x^2+x+1$. Then $w^2+w+1=0$ so $w^3-1 = (w^2+w+1)(w-1) = 0$, so $w^3=1$, so $w^4=w$. Therefore $w^4+w^2+1=w+w^2+1=w^2+w+1=0$, so $w^2$ is also a root of $x^2+x+1$.