factors of $(x+y)^p-x^p-y^p$ Assume $x,y\in\mathbb{N}$ and $p$ - prime, $p\ge 5$.
$5xy(x+y)(x^2+xy+y^2)$ divides $(x+y)^5-x^5-y^5$ (in fact, there is equality).
$7xy(x+y)(x^2+xy+y^2)^2$ divides $(x+y)^7-x^7-y^7$ (there is equality also).
$11xy(x+y)(x^2+xy+y^2)$ divides $(x+y)^{11}-x^{11}-y^{11}$ but $11xy(x+y)(x^2+xy+y^2)^2$ does not.

It is easy to prove that $pxy(x+y)$ divides $(x+y)^p-x^p-y^p$.
Does $pxy(x+y)(x^2+xy+y^2)$ divide $(x+y)^p-x^p-y^p$ always?
Is there a characterisation of $p$'s for which $pxy(x+y)(x^2+xy+y^2)^2$ divides $(x+y)^p-x^p-y^p$ ?

Edit:
For $p=17$ there is no divisibility by $(x^2+xy+y^2)^2$, similarily as for $p=11$.

Solving for $x$ in $x^2+xy+y^2=0$, we obtain
$$x = \frac{-y \pm \sqrt{y^2-4y^2}}{2}=\frac{-1 \pm i\sqrt{3}}{2}y$$
Call one solution (the one with positive imaginary part) $\xi y$, the other $\xi'y$, so that $\xi^3 = \xi'^3 = 1$. Then it's easy to show that $x = \xi y$ and $x = \xi'y$ are roots of $f(x,y)=(x+y)^p-x^p-y^p$, and so $(x-\xi y)(x-\xi'y)=x^2+xy+y^2$ divides $f$. 
$(x^2+xy+y^2)^2$ will divide $f$ precisely when $x = \xi y, \xi' y$ are multiple roots of $f$, i.e. when they are roots of $\frac{\partial f}{\partial x} = p((x+y)^{p-1}-x^{p-1})$. Letting $x=\xi y$, we have
$$0 = (\xi y + y)^{p-1}-(\xi y)^{p-1}$$
$$0 = (1+\xi)^{p-1}-\xi^{p-1}$$
$$\xi^{p-1}=(1+\xi)^{p-1}$$
$$\xi = \zeta_{p-1}(1+\xi)$$
($\zeta_{p-1}$ is some $(p-1)^{th}$ root of unity, not necessarily primitive)
$$\zeta_{p-1} = \frac{\xi}{1+\xi} = \frac{1}{2}+\frac{\sqrt{3}}{2}i$$
That last number is a complex cube root of $-1$, and a primitive cube root of $6$. Had we used $\xi'$ here, we would have gotten the other complex cube root of $-1$. In order for $\zeta_{p-1}$ to be a primitive sixth roots of $1$, we need $p-1$ to be a multiple of six, and so $p \equiv 1$(mod $6$). This is also sufficient. 
