3. [4 points] Solve the inequality $\left(\sqrt{x^{3}+2 x-58}+5\right)\left|x^{3}-7 x^{2}+13 x-3\right| \leqslant 0$.
Answer: $2+\sqrt{3}$.

Solution. The first factor on the left side is positive for any valid value of $x$. Since the second factor is always non-negative, the inequality on the domain of definition (i.e., under the condition $x^{3}+2 x-58 \geqslant 0$) is equivalent to the equation $x^{3}-7 x^{2}+13 x-3=0$. One of the integer roots of the equation is $x=3$ (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 $x-3$, we factorize it: $(x-3)\left(x^{2}-4 x+1\right)=0$. Hence, $x=3$ or $x=2 \pm \sqrt{3}$.

Substituting the obtained values of $x$ into the inequality $x^{3}+2 x-58 \geqslant 0$ for verification:

$$
\begin{aligned}
& x=3 \Rightarrow 27+6-58 \geqslant 0-\text { false; } \\
& x=2+\sqrt{3} \Rightarrow(2+\sqrt{3})^{3}+2(2+\sqrt{3})-58 \geqslant 0 \Leftrightarrow 17 \sqrt{3}-28 \geqslant 0-\text { true; } \\
& x=2-\sqrt{3} \Rightarrow(2-\sqrt{3})^{3}+2(2-\sqrt{3})-58 \geqslant 0 \Leftrightarrow-17 \sqrt{3}-28 \geqslant 0-\text { false. }
\end{aligned}
$$

Therefore, the only value that fits is $x=2+\sqrt{3}$.