3. Let $a, b$ be any two distinct positive integers. Then the minimum value of $\left|a b\left(a^{4}-b^{4}\right)\right|$ is $\qquad$ .


3.30.
$$
\begin{array}{l}
\text { Let } A=a b\left(a^{4}-b^{4}\right) \\
=a b(a-b)(a+b)\left(a^{2}+b^{2}\right) .
\end{array}
$$

First, prove: $2 \mid A$.
If $a$ and $b$ are both odd or both even, then $2 \mid (a-b)$, so $2 \mid A$;
If $a$ and $b$ are of different parity, then one of $a$ or $b$ is even, so $2 \mid ab$, hence $2 \mid A$.
Next, prove: $3 \mid A$.
If one of $a$ or $b$ is a multiple of 3, then $3 \mid ab$, so $3 \mid A$;
If neither $a$ nor $b$ is a multiple of 3, then
$a \equiv b \pmod{3}$ or $a \equiv -b \pmod{3}$.
So, $3 \mid (a-b)(a+b)$, hence $3 \mid A$.
Finally, prove: $5 \mid A$.
By $a \equiv 0, \pm 1, \pm 2 \pmod{5}$, we get
$$
a^{2} \equiv 0, \pm 1 \pmod{5} \text {. }
$$

Similarly, $b^{2} \equiv 0, \pm 1 \pmod{5}$.
Thus, $ab \equiv 0 \pmod{5}$ or $a^{2}-b^{2} \equiv 0 \pmod{5}$ or $a^{2}+b^{2} \equiv 0 \pmod{5}$.
Therefore, $5 \mid A$.
Since $2, 3, 5$ are pairwise coprime, we have
$30 \mid a b\left(a^{4}-b^{4}\right)$.
Thus, $\left|a b\left(a^{4}-b^{4}\right)\right| \geqslant 30$.
When $a=2, b=1$, $a b\left(a^{4}-b^{4}\right)=30$, so the minimum value of $\left|a b\left(a^{4}-b^{4}\right)\right|$ is 30.