<s>[INST] 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.
 [/INST] 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 $x^2+x+1$
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