prove polynomial division for any natural number Show that for any natural numbers $a$, $b$, $c~$ we have $~x^2 + x + 1|x^{3a+2} + x^{3b+1} + x^{3c}$.
Any hints on what to use?

$x^2+x+1$ splits in $\mathbf C$ as $(x-j)(x-\bar j)$, so it is enough to prove $j$ is a root of $p(x)=x^{3a+2} + x^{3b+1} + x^{3c}$. Since $j^3=1$,we have
$$p(j)=j^2+j+1=0.$$
