Problem 612

Solution:
$a^{2}+b^{2}+c^{2}=1$ and therefore $|a| \leq 1$. This implies that $a^{3} \leq |a|^{3} \leq a^{2}$, where equality holds if and only if $a=0$ or $a=1$. The same reasoning applies to $b$ and $c$. Therefore if one of the three numbers were different from 0 and from 1, we would have
$$1=a^{3}+b^{3}+c^{3}<a^{2}+b^{2}+c^{2}=1,$$
which is absurd. Consequently two of the unknowns must be equal to 0 and one equal to 1. Therefore there exist three solutions, corresponding to the three possible choices of the unknown to be set equal to 1.