<s>[INST] 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?
 [/INST] 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 $y$.
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