10. What is the value of the sum of the sixth powers of the solutions of the equation $x^{6}-16 x^{4}+16 x^{2}-1=0$?
(A) 6375
(B) 6482
(C) 6515
(D) 6660
(E) 6662.
10. The answer is $\mathbf{( E )}$. We have $x^{6}-16 x^{4}+16 x^{2}-1=\left(x^{2}-1\right)\left(x^{4}-15 x^{2}+1\right)$ (all roots are real), so if $z$ is a root different from $\pm 1$, we have $z^{4}=15 z^{2}-1, z^{6}=15 z^{4}-z^{2}=224 z^{2}-15$. Let $z_{1}, \ldots, z_{4}$ be these roots, then:

$$
\sum_{j=1}^{4} z_{j}=0, \quad \sum_{1 \leq i<j \leq 4} z_{i} z_{j}=-15
$$

$$
\begin{gathered}
\sum_{i=1}^{4} z_{i}^{2}=\left(\sum_{j=1}^{4} z_{j}\right)^{2}-2 \sum_{1 \leq i<j \leq 4} z_{i} z_{j}=30, \\
\sum_{i=1}^{4} z_{i}^{6}=224 \sum_{i=1}^{4} z_{i}^{2}-60=6660,
\end{gathered}
$$

reintroducing now the roots $\pm 1$ we have that the correct answer is 6662. (D'Aurizio)