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

Find the values of $a$, $b$, and $c$.
(Supplied by Yang Xiaoming)
2. From the given conditions, we have
$$
\begin{array}{l}
a^{2} \mid \left(a^{3}+b^{3}+c^{3}\right), b^{2} \mid \left(a^{3}+b^{3}+c^{3}\right), \\
c^{2} \mid \left(a^{3}+b^{3}+c^{3}\right).
\end{array}
$$

Since \(a\), \(b\), and \(c\) 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 \(c \geqslant 2\), \(b \leqslant 1\), which contradicts \(b \geqslant c\). Therefore, \(c=1\).
Clearly, \((1,1,1)\) is a solution.
When \(b \geqslant 2\), \(a > b > c\).
From \(a^{2} b^{2} \mid \left(a^{3}+b^{3}+1\right)\),
$$
\begin{aligned}
\Rightarrow 2 a^{3} & > a^{3}+b^{3}+1 \geqslant a^{2} b^{2} \\
& \Rightarrow a > \frac{b^{2}}{2}.
\end{aligned}
$$

Also, from \(a^{2} \mid \left(b^{3}+1\right)\),
$$
\begin{array}{l}
\Rightarrow b^{3}+1 \geqslant a^{2} > \frac{b^{4}}{4} \\
\Rightarrow 4 b^{3}+4 > b^{4}.
\end{array}
$$

When \(b \geqslant 5\), there are no solutions.
By checking \(b=2,3,4\), we find \(b=2, a=3\). Therefore, the positive integer solutions \((a, b, c)\) are
$$
\begin{array}{l}
(1,1,1),(1,2,3),(1,3,2),(2,1,3), \\
(2,3,1),(3,2,1),(3,1,2).
\end{array}
$$