8. A. Given positive integers $a$, $b$, $c$ satisfy
$$
\begin{array}{l}
a+b^{2}-2 c-2=0, \\
3 a^{2}-8 b+c=0 .
\end{array}
$$

Then the maximum value of $a b c$ is $\qquad$
8. A. 2013.

The two equations are simplified and rearranged to get
$$
(b-8)^{2}+6 a^{2}+a=66 \text {. }
$$

Given that $a$ is a positive integer and $6 a^{2}+a \leqslant 66$, we have $1 \leqslant a \leqslant 3$. If $a=1$, then $(b-8)^{2}=59$, which has no positive integer solutions; if $a=2$, then $(b-8)^{2}=40$, which has no positive integer solutions; if $a=3$, then $(b-8)^{2}=9$. Solving this, we get $b=11$ or 5. (1) If $b=11$, then $c=61$, thus, $a b c=3 \times 11 \times 61=2013$;
(2) If $b=5$, then $c=13$, thus, $a b c=3 \times 5 \times 13=195$.
In summary, the maximum value of $a b c$ is 2013.