1. 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 $1993+5 a^{2}$ $+9 a^{4}=$ $\qquad$
1. According to Vieta's formulas, we have $x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3}=75$. Thus, $a^{2}=25, a^{4}=625$. Therefore, $1993+5 a^{2}+9 a^{4}=7743$.