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.

Generally speaking , a polynomial $P \in \mathbb{C}[t]$ of degree $n$ divides some  other element of $\mathbb{C}[t]$ say Q of degree $\ge n$, if and only if
$$Q(\xi_j)=0$$ where $\{\xi_{j}: j=1, \cdots ,n\}$ is the zero set of $P$
Therefore , in this case we've $q(x)=(x-\omega)(x-\omega^2)$ and $q(x) \mid P_n(x)$, iff
$P_n (\omega^j)=0$ for $j=1,2$
Now
$$P_n(\omega)=(\omega +1)^{2n+1}+(\omega)^{n+2}=-\omega^{4n+2}+\omega^{n+2}=0$$
as $4n+2=n+2$ in $\mathbb{Z}_3$( or just as we're working in $\mathbb{Z}_{3}$)
Similarly
$$P(\omega^2)=(\omega^2 +1)^{2n+1}+(\omega^2)^{n+2}=-\omega^{2n+1}+\omega^{2n+4}=0$$
Again because we're working with the exponents in $\mathbb{Z}_{3}$
Conclusion:
$q(x) \mid P_n(x)$
