8. A three-digit number $a b c$ satisfies $a b c=a+b^{2}+c^{3}$. Then the number of three-digit numbers $\overline{a b c}$ that satisfy the condition is $\qquad$ .
8. 4 .
$$
\begin{array}{l} 
\text { Given } \\
\overline{a b c}=a+b^{2}+c^{3} \\
\Leftrightarrow 100 a+10 b+c=a+b^{2}+c^{3} \\
\Leftrightarrow(c-1) c(c+1)=99 a+b(10-b) .
\end{array}
$$

Therefore, $99 a \leqslant(c-1) c(c+1) \leqslant 99 a+25$.
When $a=1$,
$$
99 \leqslant(c-1) c(c+1) \leqslant 124 \Rightarrow c=5 \text {. }
$$

Substituting into equation (1) gives
$$
(b-3)(b-7)=0 \Rightarrow b=3,7 \text {. }
$$

Thus, in this case, we have $135, 175$.
When $a=2$,
$$
198 \leqslant(c-1) c(c+1) \leqslant 223 \Rightarrow c=6 \text {. }
$$

And $b(10-b)=12$ has no integer solutions.
Similarly, when $a=3,4,6,7,8,9$, $c$ does not exist.
When $a=5$,
$$
495 \leqslant(c-1) c(c+1) \leqslant 520 \Rightarrow c=8 \text {. }
$$

Substituting into equation (1) gives
$$
(b-1)(b-9)=0 \Rightarrow b=1,9 \text {. }
$$

Thus, in this case, we have $518, 598$.
In summary, the three-digit numbers that satisfy the condition are $135, 175$,
$518, 598$.