Prove $(x^2 + xy + y^2)^2$ divides $(x + y)^{6n + 1} - x^{6n + 1} - y^{6n + 1}$. As stated in the title, I want to show the (bivariate) polynomial $g(x, y) = (x^2 + xy + y^2)^2$ divides the polynomial $f(x, y) = (x + y)^{6n + 1} - x^{6n + 1} - y^{6n + 1}$, where $n \geq 0$.
Naturally, I resorted induction. For $n = 1$, $f(x, y) = 7xy(x + y)g(x, y)$. From $n$ to $n + 1$,
I am only able to show that $x^2 + xy + y^2$ is a factor of $f(x, y)$. Is there any trick I am missing?
 A: The key insight is that suffices to take $y=1$ because the polynomials are homogeneous. The precise details are below.
Write $z=x/y$.
Then $x^2 + xy + y^2 = y^2(z^2+z+1)$ and $(x + y)^{6n + 1} - x^{6n + 1} - y^{6n + 1}=y^{6n + 1}((z + 1)^{6n + 1} - z^{6n + 1} - 1)$.
Thus, it suffices to prove that $(z^2+z+1)^2$ divides $(z + 1)^{6n + 1} - z^{6n + 1} - 1$.
Let $h(z)=z^2+z+1$.
We have $(z+1)^{6n} \equiv 1 \bmod h(z)$.
because $(z+1)^6=(z^4 + 5 z^3 + 9 z^2 + 6 z)h(z)+1$.
We have $z^{6n} \equiv 1 \bmod h(z)$
because $z^3= (z-1)h(z)+1$.
Therefore,
$f(z)=(z + 1)^{6n + 1} - z^{6n + 1} - 1 \equiv (z+1) -z -1 = 0 \bmod h(z)$
Now
$f'(z)=(6n+1)((z + 1)^{6n} - z^{6n}) \equiv (6n+1)(1-1) = 0\bmod h(z)$
Therefore, $h$ divides both $f$ and $f'$, and so $h^2$ divides $f$.
All this can be done exactly the same without introducing $z$ but it is easy only in hindsight.
A: Let $ y = k $ = constant and let $f(x)= (x+k)^{6n+1} - x^{6n+1}-k^{6n+1}$. We know that
$x^2 + xk + k^2= (x-kw)(x-kw^2)$ where $w$, $w^2$  are cube roots of unity. If we put $x=kw $ and $kw^2$ in $f(x)$, it gives the value $0$. Therefore $(x-kw)(x-kw^2)$ divides $f(x)$ .  $\frac{df(x)}{dx}$ = $(6n+1)((x+k)^{6n} -x^{6n})$. Putting $x=kw $ and $kw^2$ in this expression we also get $0$. Therefore  $(x-kw)(x-kw^2)$ divides both $f(x)$ and $f'(x)$. Therefore ${((x-kw)(x-kw^2))}^2$ divides $f(x)$ that is $(x^2 + xk + k^2)^2$ divides $f(x)$  . By varying $k$ we get that it is true for all values of $y$.
