3. [4 points] Solve the inequality $\left(\sqrt{x^{3}-10 x+7}+1\right) \cdot\left|x^{3}-18 x+28\right| \leqslant 0$.
Answer: $x=-1+\sqrt{15}$.

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 $x^{3}-10 x+7 \geqslant 0$) is equivalent to the equation $x^{3}-18 x+28=0$. 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: $(x-2)\left(x^{2}+2 x-14\right)=0$. Hence, $x=2$ or $x=-1 \pm \sqrt{15}$.

Substituting the obtained values of $x$ into the inequality $x^{3}-10 x+7 \geqslant 0$ for verification:

$$
\begin{aligned}
& x=2 \Rightarrow 8-20+7 \geqslant 0-\text { false; } \\
& x=-1+\sqrt{15} \Rightarrow(-1+\sqrt{15})^{3}-10(-1+\sqrt{15})+7 \geqslant 0 \Leftrightarrow 8 \sqrt{15}-29 \geqslant 0-\text { true; } \\
& x=-1-\sqrt{15} \Rightarrow(-1-\sqrt{15})^{3}-10(-1-\sqrt{15})+7 \geqslant 0 \Leftrightarrow-8 \sqrt{15}-29 \geqslant 0-\text { false. }
\end{aligned}
$$

Therefore, the only value that fits is $x=-1+\sqrt{15}$.