Divisibility of polynomial 
Prove that: 
  $(x^2+x+1) \mid (x^{6n+2}+x^{3n+1}+1) $ and 
  $(x^2+x+1) \mid (x^{6n+4}+x^{3n+2}+1) $.

I saw proof in book with third roots of unity but i didn't understand it, so i want to see other solution but i dont have idea for that.
 A: Hint $\ {\rm mod}\ x^2\!+\!x\!+\!1\!:\,\ 0\equiv (x^2\!+\!x\!+\!1)(x\!-\!1) \equiv x^3-1\,\Rightarrow\,\color{#c00}{x^3} \equiv 1\,\Rightarrow\, \color{#c00}x^{\color{#c00}3n}\equiv 1\,\Rightarrow\,x^{6n}\equiv 1$
Substituting in both $\ x^{3n},x^{6n}\equiv 1\,$  shows they're $\,\equiv x^2\!+\!x\!+\!1\equiv 0,\ $ using also $\,x^4=x\cdot \color{#c00}{x^3}\equiv x\,$ 
Remark $ $ Why does integer congruence arithmetic ("modular" arithmetic) extend to polynomials? Simply because the proofs of the basic congruence laws (sum and product rules) work universally, i.e. they use only basic ring laws (distributive,commutative,associative), so the proofs remain valid in any commutative ring. This universality becomes clearer when one studies abstract algebra, which studies more structural versions of these ideas (ideals & quotient rings, a.k.a. residue rings).
A: $x^{6n+2}+x^{3n+1}+1 = (x^2+x+1) + x^2(x^{6n}-1) + x(x^{3n}-1)$
We know, $x^3-1\mid x^{3n}-1$ and $x^3-1 \mid x^{6n} - 1$ and $x^2+x+1 \mid x^3-1$ which proves the claim.
As for the second part note that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.
A: Although you have asked for another method, but I would still do the method the question wanted you to do to complement the answers already posted.
 Consider the equation $x^3=1$
So $x^3-1=0$
Which gives us $(x-1)(x^2+x+1)=0$
If $x^2+x+1$ is factorized, we get it's roots as $\frac{-1+\sqrt3i}{2}$ and $\frac{-1-\sqrt3i}{2}$, which are called $\omega$ and ${\omega}^2$. That is why we say 1,$\omega$ and ${\omega}^2$ are the cube roots of unity.
Putting $\omega$ and ${\omega}^2$ in our original equation we know that ${\omega}^3=1$ and $\omega + {\omega}^2 +1=0$.
Put $\omega$ and ${\omega}^2$ in your above question. Makes the question a breeze.
