$10 \cdot 42$ Find all three-digit numbers $\overline{a b c}$ that satisfy $\overline{a b c}=(a+b+c)^{3}$.
(China Shanghai Senior High School Mathematics Competition, 1988)
[Solution] Since $100 \leqslant \overline{a b c} \leqslant 999$, then

thus
$$
\begin{array}{c}
100 \leqslant(a+b+c)^{3} \leqslant 999, \\
5 \leqslant a+b+c \leqslant 9 .
\end{array}
$$

When $a+b+c=5$,
$$
5^{3}=125 \neq(1+2+5)^{3} .
$$

When $a+b+c=6$,
$$
6^{3}=216 \neq(2+1+6)^{3} .
$$

When $a+b+c=7$,
$$
7^{3}=343 \neq(3+4+3)^{3} .
$$

When $a+b+c=8$,
$$
8^{3}=512=(5+1+2)^{3} .
$$

When $a+b+c=9$,
$$
9^{3}=729 \neq(7+2+9)^{3} .
$$

Therefore, the required three-digit number is only 512.