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

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# Solution.

The inequality $\sqrt[2022]{x^{3}-x-\frac{1}{x}+\frac{1}{x^{3}}+3} \leq 0$ has a solution only if $x^{3}-x-\frac{1}{x}+\frac{1}{x^{3}}+3=0$. Let $t=x+\frac{1}{x}$, then $t^{3}-4 t+3=0, t^{3}-t-3 t+3=0,(t-1)\left(t^{2}+t-3\right)=0, t=1$ or $t=\frac{-1 \pm \sqrt{13}}{2}$.

The equations $x+\frac{1}{x}=1$ and $x+\frac{1}{x}=\frac{-1+\sqrt{13}}{2}$ have no solutions.

$$
\begin{gathered}
x+\frac{1}{x}=\frac{-1-\sqrt{13}}{2} \\
x^{2}+\frac{1+\sqrt{13}}{2} x+1=0 \\
D=\left(\frac{1+\sqrt{13}}{2}\right)^{2}-4=\frac{2 \sqrt{13}-2}{4}=\left(\frac{\sqrt{2 \sqrt{13}-2}}{2}\right)^{2} \\
x_{1,2}=\frac{-1-\sqrt{13} \pm \sqrt{2 \sqrt{13}-2}}{4}
\end{gathered}
$$

$$
\text { Answer. } \frac{-1-\sqrt{13} \pm \sqrt{2 \sqrt{13}-2}}{4} \text {. }
$$