Problem 5. (20 points) Solve the inequality $\sqrt[2022]{x^{3}-4 x-\frac{4}{x}+\frac{1}{x^{3}}+6} \leq 0$.


Solution.

The inequality $\sqrt[2022]{x^{3}-4 x-\frac{4}{x}+\frac{1}{x^{3}}+6} \leq 0$ has a solution only if $x^{3}-4 x-\frac{4}{x}+\frac{1}{x^{3}}+6=0$.

Let $t=x+\frac{1}{x}$, then $t^{3}-7 t+6=0, t^{3}-t-6 t+6=0,(t-1)\left(t^{2}+t-6\right)=0, t=1$, or $t=2$, or $t=-3$.

The equation $x+\frac{1}{x}=1$ has no roots.

$$
\begin{gathered}
x+\frac{1}{x}=2, x^{2}-2 x+1=0,(x-1)^{2}=0, x=1 \\
x+\frac{1}{x}=-3, x^{2}+3 x+1=0, D=(\sqrt{5})^{2}, x_{1,2}=\frac{-3 \pm \sqrt{5}}{2}
\end{gathered}
$$

$$
\text { Answer. }\left\{1 ; \frac{-3 \pm \sqrt{5}}{2}\right\} \text {. }
$$