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.

Induction is really helpful here, did you try it?
lets check the case $n=0$:
$$(x+1)^{2\cdot0+1}+x^{0+2} = x+1+x^2$$ which is of course multiplier of $x^2+x+1$.
Assume, that your claim is true for every $k\leq n$, and lets prove the next step ($n+1$):
we need to prove that $x^2+x+1$ is factor of $(x+1)^{2(n+1)+1}+x^{n+1+2}$
and this can be proved easily:
$$\begin{align}(x+1)^{2n+3}+x^{n+3}&=(x+1)^2\cdot (x+1)^{2n+1}+x\cdot x^{n+2}\\
&=(x^2+2x+1)\cdot(x+1)^{2n+1}+x\cdot x^{n+2}\\
&=(x^2+x+1)\cdot(x+1)^{2n+1}+x\cdot(x+1)^{2n+1}+x\cdot x^{n+2}\\
&=(x^2+x+1)\cdot(x+1)^{2n+1}+x\cdot((x+1)^{2n+1}+x^{n+2})
\end{align}$$
where both terms are multipliers of $x^2+x+1$
