One of $a^5b-ab^5,b^5c-bc^5,c^5a-ca^5$ is divisible by $8$ if $a,b,c∈\mathbb{Z}^+$ Given that $a,b,c$ are there distinct positive integers, prove that among $a^5b-ab^5,b^5c-bc^5,c^5a-ca^5$, there is at least one that is divisible by $8$.
I have no clue what to do and cannot even create cases as these are three random whole numbers. What do I do? Please help me guys!!!
Even a hint is graciously accepted!!!

The three numbers are cyclic permutations of $ab(a^4-b^4)$.  Therefore, at least one of the products must be formed from either $2$ even numbers or $2$ odd numbers.
If $a$ and $b$ are both even, then so is $(a^4-b^4)$, so $ab(a^4-b^4)$ is the product of three even numbers, which must be divisible by $8$.
If $a$ and $b$ are both odd, then $a^4 \equiv b^4 \pmod{8}$ (because any odd number is a square root of $1$ (mod $8$)), so $8|(a^4-b^4)$ and therefore $8|ab(a^4-b^4).$
