Prove that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$ 
Let $n,x,y,z$ be positive integers such that $$(x-y)^2+(y-z)^2+(z-x)^2 = (2n+1)xyz.$$ Prove that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$.

The original equation doesn't have $x^3,y^3,z^3$, so how can we show that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$? Expanding the given equation seems to make the question more complicated, and also how do we get the $6$?

This is linked to a well-known identity (a standard exercise in high-school):
\begin{align}x^3+y^3+z^3-3xyz&=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\&=\tfrac12(x+y+z)\bigl((x-y)^2+(y-z)^2+(z-x)^2\bigr)\end{align}
