Divisibility Number Theory problem, explanation needed

Question

I can't understand the solution of the following problem: $x$,$y$,$z$ are pairwise distinct natural numbers show that $(x-y)^5$ + $(y-z)^5$ + $(z-x)^5$ is divisible by $5(x-y)(y-z)(z-x)$. No need to explain the div. by 5.

The sol. says: $(x-y)^5$ + $(y-z)^5$ + $(z-x)^5$ is $zero$ for $x=y$, $y=z$, $z=x$. So the terms $(x-y)$, $(y-z)$, $(z-x)$ can be factored out.

This is the 106th problem chap. 6 form "Problem solving strategies" by A. Engel If you have alternative solution pls feel free to post it.

Answer 1

For example, the given expression is divisible by $x-y$ if and only if $y$ is a root of the polynomial regarded as a function of $x$. But substituting $x=y$ indeed gives zero, so that $x-y$ must be a factor.

Answer 2

Let $x-y=a$ and $y-z=b$.

Thus, $z-x=-(a+b)$ and $$\sum_{cyc}(x-y)^5=a^5+b^5-(a+b)^5=-5a^4b-10a^3b^2-10a^2b^3-5ab^4=$$ $$=-5ab(a^3+2a^2b+2ab^2+b^3)=-5ab(a+b)(a^2-ab+b^2+2ab)$$ and we are done!