Prove $30|(a^3b-ab^3) $

Hints:

$$a^3b-ab^3=ab(a^2-b^2)=ab(a-b)(a+b)$$

Hints: check that no matter what the parity of $\;a,b\;$ is, the above product is always even and thus divisible by two.

Also, no matter what $\;a,\,b\,$ are, the product is always divisible by $\;3\;$ .

I'll leave to you the funniest part: to prove the product of two of the three different integers chosen is always divisible by $\;5\;$ (not necessarily a given pair $\;a,b\;$ !) . Observe that you can reduce this to the case where the three different integers are different $\;\pmod5\;$ , otherwise we're done (why?) .

By Fermat's little theorem (or simply by direct calculation), $x^2\equiv x$ mod $2$ and $x^3\equiv x$ mod $3$ for any integer $x$. It follows that

$$a^3b\equiv ab\equiv ab^3\mod 6$$

for any pair of integers $a,b$, hence $6$ divides $a^3b-ab^3$ for any pair. It remains to show that $5$ divides $a^3b-ab^3=ab(a^2-b^2)$ for at least one of the three pairs. But there are only three squares mod $5$, namely $0$, $1$, and $-1$. So given three integers, either two of their squares, say $a^2$ and $b^2$ belong to the same residue class, in which case $5$ divides $a^2-b^2$, or one of the three squares, say $a^2$ is $0$ mod $5$, in which case $5$ divides $a$ (and you can let $b$ be either of the other two numbers). In either case $5$ divides $ab(a^2-b^2)=a^3b-ab^3$ for the chosen pair, and we are done.