8 . All three-digit numbers $a b c$ that satisfy $a b c=(a+b+c)^{3}$ are $\qquad$
8. $\because 100 \leqslant a b c=(a+b+c)^{3} \leqslant 999$,
$\therefore 5<a+b+c$ S 9 . Directly calculating $5^{3}=125$, $6^{3}=216,7^{3}=343,8^{3}=512,9^{3}=729$, we can see that only the three-digit number 512 meets the conditions of the problem.