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.

$x^{6n+2}+x^{3n+1}+1 = (x^2+x+1) + x^2(x^{6n}-1) + x(x^{3n}-1)$
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 $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.
