1. Solve the equation

$$
(\sqrt{3-\sqrt{8}})^{x}+(\sqrt{3+\sqrt{8}})^{x}=6
$$
Solution. If in the equation we replace $\sqrt{8-\sqrt{3}}$ with $\frac{1}{\sqrt{3+\sqrt{8}}}$, we will get an equivalent equation

$$
(\sqrt{3+\sqrt{8}})^{2 x}-6(\sqrt{3+\sqrt{8}})^{x}+1=0
$$

If we let $(\sqrt{3+\sqrt{8}})^{x}=y$, we will get the following quadratic equation

$$
y^{2}-6 y+1=0
$$

whose solutions are $y=3+2 \sqrt{2}$ and $y=3-2 \sqrt{2}$. From $(\sqrt{3+\sqrt{8}})^{x}=3+2 \sqrt{2}$, we get $x=2$, and from $(\sqrt{3+\sqrt{8}})^{x}=\frac{1}{3+2 \sqrt{2}}=(3+2 \sqrt{2})^{-1}$, we get $x=-2$.

Therefore, the equation has two solutions $x_{1}=2$ and $x_{2}=-2$.