Problem with polynomial divisilibity proof. 
Show that for any natural $k$, $m$, $n$ polynomial $x^{3k}+x^{3m+1}+x^{3m+2}$ is divisible by $x^2+x+1$.

My observation:
$$x^a=(x^{a-2}-x^{a-3})(x^2+x+1)+x^{a-3}$$
so,
$$x^{3k}+x^{3m+1}+x^{3m+2}=(x^{3k-2}-x^{3k-3})(x^2+x+1)+x^{3k-3}+(x^{3m-1}-x^{3m-2})(x^2+x+1)+x^{3m-2}+(x^{3n}-x^{3n-1})(x^2+x+1)+x^{3n-1}$$
We can factor $x^2+x+1$, but i don't see any conclusion after that.
 A: Note that one way is to show that the roots of $x^2+x+1$ satisfy $x^{3k}+x^{3m+1}+x^{3n+2}=0$, as suggested in comments. Here is an easy approach without using complex numbers:
$$x^{3k}+x^{3m+1}+x^{3n+2}=x^{3k}-1+x(x^{3m}-1)+x^2(x^{3n}-1)+x^2+x+1$$
Now note that $$x^2+x+1\bigg|x^3-1\bigg|x^{3a}-1\ \ \forall a\in\Bbb N$$
Therefore, RHS of the first equation is divisible by $x^2+x+1$ and we are done.
A: Observe that $q(x)=x^2+x+1$ divides $p(x)=x^{3k}+x^{3m+1}+x^{3n+2}$ if and only if the roots of $q(x)$ are roots of $p(x)$.
Since $x^3-1 = (x-1)(x^2+x+1)$, the roots of $q(x)$ are $\alpha$ and $\beta$ and they are the primitive cubic roots of unity and obviously the following holds:
\begin{gather}
\alpha^3 = \beta^3 =1\\
\alpha^2+\alpha+1 = \beta^2+\beta+1=0
\end{gather}
Now we simply have:
$$
p(\alpha)=\alpha^{3k}+\alpha^{3m+1}+\alpha^{3n+2} = (\alpha^{3})^k+\alpha(\alpha^{3})^m+\alpha^2(\alpha^{3})^n =1+ \alpha+ \alpha^2 =0
$$
and the same thing happens to $\beta$. So $p(\alpha)=p(\beta)=0$ and $x^2+x+1\ | \ p(x)$ for all $n,m,k$
