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 $6$?

We're given that
$$(x-y)^2+(y-z)^2+(z-x)^2=(2n+1)xyz$$
Hence
$$\frac{(x-y)^2+(y-z)^2+(z-x)^2}{xyz}=(2n+1)$$
Substitute this into the expression given:
$$
\begin{aligned}
(2n+1)(x+y+z)+6
=
\frac{(x-y)^2+(y-z)^2+(z-x)^2}{xyz}(x+y+z)+\frac{6xyz}{xyz}
\end{aligned}
$$
After expanding and simplifying, you will get to
$$\frac{2(x^3+y^3+z^3)}{xyz}$$
Which certainly does divide $x^3+y^3+z^3$ if one of $x,y,z$  is even...
