If $x$ satisfies $x^2+x+1=0$, then the value of $x^{100}+x^{200}$ is:

Multiply the given equation by $(x-1):(x-1)(x^2+x+1)=x^3-1=0$, so $x^3=1$ while $x$ ≠ 1. Hence $x^{100}=x^{99}\cdot x=x$ and $x^{200}=x^{198}\cdot x^2=x^2$, so the sum is $x+x^2$, which the original equation says equals −1. (2 assumes $x^3=1$ makes every power of $x$ equal 1; 0 is the value of $1+x+x^2$, one term more than asked.) The roots are the non-real cube roots of 1, and ω + ω² = −1.