2. Find all triples of positive integers $(a, b, c)$ such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$.
2. $(a, b, c)=(1,2,3)$ and all its permutations. Since $a, b, c$ are symmetric, we can assume $a \geqslant b \geqslant c$.
From $3 a^{3} \geqslant a^{3}+b^{3}+c^{3}>a^{3}$
$\Rightarrow 3 a^{3} \geqslant(a b c)^{2}>a^{3}$
$\Rightarrow 3 a \geqslant(b c)^{2}>a$.
Also, $b^{3}+c^{3}=a^{2}\left(b^{2} c^{2}-a\right)>a^{2}$, then
$18 b^{3} \geqslant 9\left(b^{3}+c^{3}\right)>9 a^{2} \geqslant b^{4} c^{4} \geqslant b^{3} c^{5}$
$\Rightarrow 18>c^{5} \Rightarrow c=1$.
Thus, $a^{3}+b^{3}+1=(a b)^{2}$.
If $a=b$, then $2 b^{3}+1=b^{4}$, there is no such positive integer $b$.
If $a>b$, then $a^{3}-b^{3} \geqslant(b+1)^{3}-b^{3}>1$.
Therefore, $2 a^{3}>a^{3}+b^{3}+1>a^{3}$.
This implies, $2 a^{3}>a^{2} b^{2}>a^{3} \Rightarrow 2 a>b^{2}>a$.
From $4\left(1+b^{3}\right)=4 a^{2}\left(b^{2}-a\right) \geqslant 4 a^{2}>b^{4}$, we get
$4>b^{3}(b-4)$.
Thus, $2 \leqslant b \leqslant 4$.
If $b=2$, then $a^{3}-4 a^{2}+9=0 \Rightarrow a=3$;
If $b=3$, then $a^{3}-9 a^{2}+28=0$, there is no positive integer $a>3$ satisfying this equation;

If $b=4$, then $a^{3}-16 a^{2}+65=0$, there is no positive integer $a>4$ satisfying this equation.
In conclusion, $(a, b, c)=(1,2,3)$ and all its permutations.