3. [4 points] Solve the equation $(x+3) \sqrt{x^{3}-x+10}=x^{2}+5 x+6$.


Answer: $2 ; \frac{\sqrt{13}-1}{2}$.

Solution. Factoring the right side, we get $(x+3) \sqrt{x^{3}-x+10}=(x+$ $3)(x+2)$. From this, there are two possibilities: either $x+3=0$ (then $x=-3$, which does not fit the domain of definition, as the expression under the root is negative), or $\sqrt{x^{3}-x+10}=x+2$. We solve the latter equation:

$$
\left\{\begin{array} { l } 
{ x ^ { 3 } - 3 x + 1 0 = ( x + 2 ) ^ { 2 } , } \\
{ x + 2 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x ^ { 3 } - x ^ { 2 } - 5 x + 6 = 0 , } \\
{ x \geqslant - 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x-2)\left(x^{2}+x-3\right)=0 \\
x \geqslant-2
\end{array}\right.\right.\right.
$$

The equation in the system has roots $x=2, x=\frac{-1 \pm \sqrt{13}}{2}$, and of these, the inequality is satisfied by $x=2$ and $x=\frac{-1+\sqrt{13}}{2}$. These are the answers to the problem.