Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder I am in my pre-academic year. We recently studied the Remainder sentence (at least that's what I think it translates) which states that any polynomial can be written as $P = Q\cdot L + R$
I am unable to solve the following:

Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder.


I am going to prove it by modulo. In modulo $$
x^{2}+x+1 \equiv 0\left(\bmod x^{2}+x+1\right) \Rightarrow x+1\equiv -x^2\textrm{ and } x^3 \equiv 1 \left(\bmod x^{2}+x+1\right).
$$
In modulo $x^{2}+x+1$, we have
$$
\begin{aligned}
(x+1)^{2 n+1}+x^{n+2} \equiv &\left(-x^{2}\right)^{2 n+1}+x^{n+2} \\
\equiv &-x^{4 n+2}+x^{n+2} \\
\equiv & x^{n+2}\left(1-x^{3 n}\right) \\
\equiv & x^{n+2}\left(1-1\right) \\
\equiv & 0
\end{aligned}
$$
Therefore it has no remainder when divided by $x^{2}+x+1$.
