Proving that $ 30 \mid ab(a^2+b^2)(a^2-b^2)$ How can I prove that $30 \mid ab(a^2+b^2)(a^2-b^2)$ without using $a,b$ congruent modulo $5$ and then 
$a,b$ congruent modulo $6$ (for example) to show respectively that $5 \mid ab(a^2+b^2)(a^2-b^2)$ and 
$6 \mid ab(a^2+b^2)(a^2-b^2)$? 
Indeed this method implies studying numerous congruences and is quite long.

Below I show it is a very special case of  following generalization of the Euler-Fermat theorem.
Theorem $\  $ Suppose that $\rm\ n\in \mathbb N\ $ has the prime factorization $\rm\:n = p_1^{e_{\:1}}\cdots\:p_k^{e_k}\ $ and suppose  that for all $\rm\,i,\,$  $\rm\ e\ge e_i\ $ and $\rm\ \phi(p_i^{e_{\:i}})\mid f.\ $ Then $\rm\ n\mid (ab)^e\,(a^f-b^f)\ $ for all $\rm\: a,b\in \mathbb Z.$ 
Proof $\ $ Notice that if $\rm\ p_i\mid ab\ $ then $\rm\:p_i^{e_{\:i}}\ |\ (ab)^e\ $ by $\rm\ e_i \le e.\: $ Else $\rm\:a\:$ and $\rm\:b\:$ are coprime to $\rm\: p_i\:$ so by Euler's phi theorem, $\rm\ mod\ q = p_i^{e_{\:i}}\!: \ a^{\phi(q)}\equiv 1\equiv b^{\phi(q)} \Rightarrow\ a^f\equiv 1\equiv b^f\: $ by $\rm\: \phi(q)\mid f.\ $  Thus since all $\rm\ p_i^{e_{\:i}}\ |\  (ab)^e\ (a^f - b^f\:)\ $ so too does their lcm = product = $\rm\: n.$
Corollary $\rm\ \ 6p\mid a\: b^p - b\: a^p\ $ for all $\rm\:a,b\in \mathbb Z,\:$ prime $\rm\:p\ne 2,3$ 
Specializing $\rm\ p = 5\ $ yields the result sought in the question.
Remark $\: $ Note that the proof of the general theorem is less work than some other direct proofs of your special case. For some other variations (e.g. Carmichael's) see my posts here.
