Example 5 Given that $a, b, c$ are pairwise coprime positive integers, and satisfy
$$
a^{2} \mid \left(b^{3}+c^{3}\right), b^{2} \mid \left(a^{3}+c^{3}\right), c^{2} \mid \left(a^{3}+b^{3}\right) \text {. }
$$

Find the values of $a, b, c$.
(Eighth China Southeast Mathematical Olympiad)
Given:
$a^{2} \backslash\left(a^{3}+b^{3}+c^{3}\right), b^{2} \backslash\left(a^{3}+b^{3}+c^{3}\right)$,
$c^{2} \mid\left(a^{3}+b^{3}+c^{3}\right)$.
Since $a, b, c$ are pairwise coprime, we have $a^{2} b^{2} c^{2} \mid\left(a^{3}+b^{3}+c^{3}\right)$.
Assume without loss of generality that $a \geqslant b \geqslant c$.
Then $3 a^{3} \geqslant a^{3}+b^{3}+c^{3} \geqslant a^{2} b^{2} c^{2} \Rightarrow a \geqslant \frac{b^{2} c^{2}}{3}$.
Also, $2 b^{3} \geqslant b^{3}+c^{3} \geqslant a^{2}$
$\Rightarrow 2 b^{3} \geqslant \frac{b^{4} c^{4}}{9} \Rightarrow b \leqslant \frac{18}{c^{4}}$.
When $c \geqslant 2$, $b \leqslant 1$. This contradicts $b \geqslant c$.
Thus, $c=1$.
Clearly, $(1,1,1)$ is a solution.
When $b \geqslant 2$, $a>b>c$.
By $a^{2} b^{2} \mid \left(a^{3}+b^{3}+1\right)$
$\Rightarrow 2 a^{3} \geqslant a^{3}+b^{3}+1 \geqslant a^{2} b^{2}$
$\Rightarrow a \geqslant \frac{b^{2}}{2}$.
Also, by $a^{2} \mid \left(b^{3}+1\right) \Rightarrow b^{3}+1 \geqslant a^{2} \geqslant \frac{b^{4}}{4}$.
When $b>5$, there are no solutions.
Upon verification, $b=2, c=3$.
In summary, the positive integer solutions $(a, b, c)$ are $(1,1,1),(1,2,3)$ and their permutations.