We have $a^3+7=8\pmod{9}$ and so $a=1,4,7\pmod{9}$. Clearly we must have $a\le 12$ so $a$ is equal to one of $1,4,7,10$ (and this gives a maximum for $b$ and $c$ in each of these cases). But $1^3+9\cdot 1^2+9\cdot 1+7<4^3+9\cdot 4^2+9\cdot 4+7<7^3+9\cdot 7^2+9\cdot 7+7<1997$ so it is impossible if $a=1,4,7$. If $a=10$ then of course $b=c=10$ works. It is in fact the only solution since if we decrease $b$ or $c$ equality cannot be attained. So $(10,10,10)$ is the only solution.