10. Given that the real number $a$ satisfies $a^{2}-a-1=0$. Then the value of $a^{8}+7 a^{-4}$ is - $\qquad$
10.48 .

From $a^{2}-a-1=0$, we get $a-a^{-1}=1$, then, $a^{2}+a^{-2}$
$$
\begin{aligned}
=3, & a^{4}+a^{-4}=7 . \\
& a^{8}+7 a^{-4}=a^{4}\left(a^{4}+a^{-4}\right)+7 a^{-4}-1 \\
& =7\left(a^{4}+a^{-4}\right)-1=7 \times 7-1=48 .
\end{aligned}
$$