8. Given positive integers $a, b, c$ satisfy
$$
(a!)(b!)=a!+b!+c! \text {. }
$$

then $c\left(a^{5}+b^{5}+c^{2}\right)+3=$ $\qquad$
8.2011 .
$$
\begin{array}{l}
\text { Given }(a!)(b!)=a!+b!+c! \\
\Rightarrow(a!-1)(b!-1)=c!+1 \text {. }
\end{array}
$$

Assume without loss of generality that $a \geqslant b$. Clearly, $c > a$. Then $(a!-1) \mid(c!+1)$.
Also, $c!+1=\frac{c!}{a!}(a!-1)+\frac{c!}{a!}+1$, so $(a!-1) \left\lvert\,\left(\frac{c!}{a!}+1\right)\right.$
$$
\text { Hence } \frac{c!}{a!}+1 \geqslant a!-1 \Rightarrow c!\geqslant a!(a!-2) \text {. }
$$

Also, $c!=(a!-1) \backslash a!\leqslant a!(a!-2)$, combining with equation (1) we get $a=b$, and $c!=a!(a!-2)$.
Clearly, $a \geqslant 3$.
If $c \geqslant a+3$, then $\frac{c!}{a!} \equiv 0(\bmod 3)$.
But $a!-2 \equiv 1(\bmod 3)$, a contradiction.
If $c=a+2$, then
$$
\begin{array}{l}
(a+2)(a+1)=a!-2 \\
\Rightarrow a^{2}+3 a+4=a! \\
\Rightarrow a \mid 4 \Rightarrow a=4,
\end{array}
$$

which does not satisfy equation (2), a contradiction.
If $c=a+1$, then
$$
a+1=a!-2 \Rightarrow a+3=a!\Rightarrow a=3 \text {. }
$$

Thus, $a=b=3, c=4$.
In this case, $c\left(a^{5}+b^{5}+c^{2}\right)+3=2011$.