We know $\omega, \omega^{2}$ are the roots of the polynomial $z^2+z+1$ then putting $\omega$ and also $\omega^{2}$ we get that $z^{2n}+z^{n}+1$ is $0$

It is a IMO problem.If n is multiple of 3 then z6n*+z3n*+1 is divisible by z2*+z+1.But omega is root of z2*+z+1 so omega is must be root of z6n*+z3n*+1. which is not possible.since omega cube =1.So n isnot multiple of 3.

Let P(z)=(z^n - 1)(z^(2n) + z^n + 1) = z^(3n) - 1 . w,w² are roots of P , w=cis(2π/3) . But w,w² aren't roots of P'(z)=3n z^(3n-1). Then w,w² are roots of multiplicity 1 of P. Then w, w² are roots of (z^(2n) + z^n + 1) if and only if w,w² aren't roots of (z^n - 1) if and only if n is not a multiple of 3

blud what the sigma? roots of z^2+z+1 are the complex numbers w, w^2 with w^3=1. now z^2n+z^n+1 plug in w and w^2 to see that they are roots only when n is not a multiple of 3. rizz