Proving $x^2 + x + 1$ is a factor of $P_n (x)=(x+1)^{2n+1} + x^{n+2}$. Further says to consider $P_n(\omega)$ and $P_n(\omega^2)$ wherein $\omega$ is a cube root of unity. $\omega \not=1$.
Found this from a examination paper with no solutions. I understand it relates to roots of unity, but I'm unsure how I can bring that into play. Any explanation on how to approach these types of questions is appreciated thank you.

Let $\omega$ be any root of $x^2+x+1$. We will show that $\omega$ is a root of $(x+1)^{2n+1} + x^{n+2}$.
First note that we can rearrange $x^2+x+1$ to $x+1 = -x^2$ so we know that $(\omega+1)^{2n+1} = (-\omega^2)^{2n+1} = -\omega^{4n+2}$.
Thus $(\omega+1)^{2n+1} + \omega^{n+2} = -\omega^{4n+2} + \omega^{n+2} = -\omega^{n+2}(\omega^{3n} - 1).$
So we just need to show that $\omega^{3n} - 1 = 0$.
This can been shown via the identity $u^3-1 = (u-1)(u^2+u+1)$ with $u = \omega^n$.
