3. [4 points] Solve the inequality $\left(\sqrt{x^{3}+x-90}+7\right)\left|x^{3}-10 x^{2}+31 x-28\right| \leqslant 0$.
Answer: $3+\sqrt{2}$.

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 $x^{3}+x-90 \geqslant 0$) is equivalent to the equation $x^{3}-10 x^{2}+31 x-28=0$. One of the integer roots of the equation is $x=4$ (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 $x-4$, we factorize it: $(x-4)\left(x^{2}-6 x+7\right)=0$. Hence, $x=4$ or $x=3 \pm \sqrt{2}$.

Substituting the obtained values of $x$ into the inequality $x^{3}+x-90 \geqslant 0$ for verification:

$$
\begin{aligned}
& x=4 \Rightarrow 64+4-90 \geqslant 0-\text { false; } \\
& x=3-\sqrt{2} \Rightarrow(3-\sqrt{2})^{3}+(3-\sqrt{2})-90 \geqslant 0 \Leftrightarrow-30 \sqrt{2}-42 \geqslant 0-\text { false; } \\
& x=3+\sqrt{2} \Rightarrow(3+\sqrt{2})^{3}+(3+\sqrt{2})-90 \geqslant 0 \Leftrightarrow 30 \sqrt{2}-42 \geqslant 0-\text { true; }
\end{aligned}
$$

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