Example 8 The roots $x_{1}, x_{2}$ of the equation $x^{2}-a x-a=0$ satisfy the relation $x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3}=75$. Then $1993+5 a^{2}+$ $9 a^{4}=$ $\qquad$
Solution: By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=a, x_{1} x_{2}=-a . \\
\therefore x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3} \\
=\left(x_{1}+x_{2}\right)^{3}-3 x_{1} x_{2}\left(x_{1}+x_{2}\right)+\left(x_{1} x_{2}\right)^{3} \\
=a^{3}-3(-a) a+(-a)^{3}=3 a^{2} .
\end{array}
$$

According to the problem, $3 a^{2}=75$, so $a^{2}=25, a^{4}=625$.
$$
\begin{array}{l}
\therefore 1993+5 a^{2}+9 a^{4} \\
=1993+5 \times 25+9 \times 625=7743 .
\end{array}
$$