6. For what values of the parameter $a$ do the roots $x_{1}, x_{2}$ of the equation $x^{2}-2 a x-\frac{1}{a^{2}}=0$ satisfy the equality $x_{1}^{4}+x_{2}^{4}=16+8 \sqrt{2}$?
Solution. We will use Vieta's theorem: $x_{1} x_{2}=-\frac{1}{2 a^{2}}, x_{1}+x_{2}=2 a$, and express the sum of the fourth powers:

$$
\begin{aligned}
& x_{1}^{4}+x_{2}^{4}=\left(x_{1}+x_{2}\right)^{4}-4 x_{1} x_{2}\left(\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right)-6 x_{1}^{2} x_{2}^{2}=16 a^{4}-4\left(-\frac{1}{a^{2}}\right) 4 a^{2}+2\left(-\frac{1}{a^{2}}\right)^{2} \\
& =16 a^{4}+16+\frac{2}{a^{4}}=16+8 \sqrt{2} \Rightarrow 16 a^{8}-8 \sqrt{2} a^{4}+2=0 \Rightarrow\left(2 \sqrt{2} a^{4}-1\right)^{2}=0
\end{aligned}
$$

Answer: $a= \pm \sqrt[8]{1 / 8}$