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.

If $n=0$, then it is trivial.

For $n>0$, we have $$(x+1)^{2n+1}+x^{n+2}\\=(x^2+2x+1)(x+1)^{2n-1}+x^{n+2}\\=(x^2+x+1)(x+1)^{2n-1}+x(x+1)^{2n-1}+x^{n+2}\\=(x^2+x+1)(x+1)^{2n-1}+x((x+1)^{2n-1}+x^{n+1})$$. By using induction, suppose $(x+1)^{2n-1}+x^{n+1}$ can be divided by $x^2+x+1$, then, $(x+1)^{2n+1}+x^{n+2}$ also can be divided by $x^2+x+1$.

Hint: try to check if complex roots of $x^2+x+1$ are also roots for $(x+1)^{2n+1}+x^{n+2}$.

Also there is another way to prove it. Let see that $((x+1)^2-x)\cdot (x+1)^{2n-1}$ also divided by $x^2+x+1$. So we just need to prove that $x\cdot (x+1)^{2n-1}+x^{n+2}$ is divided by $x^2+x+1$. In this way we can come to prove that $(x^k\cdot (x+1)^{2n+1-2\cdot k}+x^{n+2}), \dots ,$ $(x^n(x+1)+x^{n+2})$ are all divided by $x^2+x+1$. But it's true, because $x^n(x+1)+x^{n+2} = x^n(x^2+x+1) $