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.


Suppose $a$ is a root of $x^2+x+1=0$, then we have both $$a+1=-a^2$$ and $$a^3=1$$
Let $f(x)=(x+1)^{2n+1}+x^{n+2}$ then $$f(a)=(-a^2)^{2n+1}+a^{n+2}=-a^{4n+2}+a^{n+2}=-a^{n+2}+a^{n+2}=0$$
Since the two distinct roots of the quadratic are also roots of $f(x)$ we can use the remainder theorem to conclude that the remainder is zero.
