Find all triplets of natural numbers $(a, b, c)$ such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$.
Suppose, without loss of generality, that $a \geqslant b \geqslant c \geqslant 1$. Since our equation involves terms of different degrees, we start by looking at what inequalities we might obtain. Here, we observe that:
口 $3 a^{3} \geqslant a^{3}+b^{3}+c^{3}=(a b c)^{2}$, and thus $3 a \geqslant(b c)^{2}$;

$\Delta a^{2}$ divides $(a b c)^{2}-a^{3}=b^{3}+c^{3}$, and thus $a^{2} \leqslant b^{3}+c^{3} \leqslant 2 b^{3}$.

Therefore, $b^{3} c^{5} \leqslant(b c)^{4} \leqslant 9 a^{2} \leqslant 18 b^{3}$, so $c^{5} \leqslant 18217$, and we conclude that 217 is prime, hence $a=1$;

$\triangleright$ if $b=7$, then $a^{2}$ divides $b^{3}+1=344=2^{3} \times 43$, so $a$ divides 2;

$\Delta$ if $b=8$, then $a^{2}$ divides $b^{3}+1=513$, which is not divisible by any prime $p \leqslant 19$; since $23^{2}=529>513$, we conclude that 513 is prime, hence $a=1$;

$\Delta$ if $b=9$, then $a^{2}$ divides $b^{3}+1=730=2 \times 5 \times 73$, so $a=1$.

The only two plausible triplets are $(a, b, c)=(1,1,1)$ and $(a, b, c)=(3,2,1)$. We then verify that $1^{3}+1^{3}+1^{3}=3 \neq 1=(1 \times 1 \times 1)^{2}$ and $3^{3}+2^{3}+1^{3}=26=(3 \times 2 \times 1)^{2}$, and we conclude that the only solutions are $(a, b, c)=(3,2,1)$ and its five permutations.