3. [4 points] Solve the inequality $\left(\sqrt{x^{3}-18 x-5}+2\right) \cdot\left|x^{3}-4 x^{2}-5 x+18\right| \leqslant 0$.
Answer: $x=1-\sqrt{10}$.

Solution. The first factor on the left side is positive for any admissible value of $x$. Since the second factor is always non-negative, the inequality on the domain of definition (i.e., under the condition $x^{3}-18 x-5 \geqslant 0$) is equivalent to the equation $x^{3}-4 x^{2}-5 x+18=0$. One of the integer roots of the equation is $x=2$ (it is not difficult to find by trial among the divisors of the free term of the equation). Performing the division of the left side by the polynomial $x-2$, we factorize it: $(x-2)\left(x^{2}-2 x-9\right)=0$. Hence, $x=2$ or $x=1 \pm \sqrt{10}$.

Substituting the obtained values of $x$ into the inequality $x^{3}-18 x-5 \geqslant 0$ for verification:

$$
\begin{aligned}
& x=2 \Rightarrow 8-36-5 \geqslant 0-\text { false; } \\
& x=1+\sqrt{10} \Rightarrow(1+\sqrt{10})^{3}-18(1+\sqrt{10})-5 \geqslant 0 \Leftrightarrow-5 \sqrt{10}+8 \geqslant 0-\text { false; } \\
& x=1-\sqrt{10} \Rightarrow(1-\sqrt{10})^{3}-18(1-\sqrt{10})-5 \geqslant 0 \Leftrightarrow 5 \sqrt{10}+8 \geqslant 0-\text { true. }
\end{aligned}
$$

Therefore, the only value that fits is $x=1-\sqrt{10}$.