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

Find the values of $a, b, c$. ${ }^{[1]}$
(Eighth China Southeast Mathematical Olympiad)
Given:
$$
\begin{array}{l}
a^{2} \mid \left(a^{3}+b^{3}+c^{3}\right), b^{2} \mid \left(a^{3}+b^{3}+c^{3}\right), \\
c^{2} \mid \left(a^{3}+b^{3}+c^{3}\right).
\end{array}
$$

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 \(a \geqslant b \geqslant c\).
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, \(a^{2} \mid \left(b^{3}+1\right) \Rightarrow b^{3}+1 \geqslant a^{2} \geqslant \frac{b^{4}}{4}\).
When \(b > 5\), there is no solution.
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.