3. [4 points] Solve the equation $\left(\frac{x}{2 \sqrt{2}}+\frac{5 \sqrt{2}}{2}\right) \sqrt{x^{3}-64 x+200}=x^{2}+6 x-40$.
Answer: $6, \sqrt{13}+1$.

Solution. Factoring the right-hand side and multiplying both sides by $2 \sqrt{2}$, we get $(x+10) \sqrt{x^{3}-64 x+200}=2 \sqrt{2}(x-4)(x+10)$. From here, there are two possibilities: either $x+10=0$ (then $x=-10$, which does not fit the domain of definition, as the expression under the square root is negative), or $\sqrt{x^{3}-64 x+200}=\sqrt{8}(x-4)$. We solve the latter equation:

$$
\left\{\begin{array} { l } 
{ x ^ { 3 } - 6 4 x + 2 0 0 = 8 ( x - 4 ) ^ { 2 } , } \\
{ x - 4 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x ^ { 3 } - 8 x ^ { 2 } + 7 2 = 0 , } \\
{ x \geqslant 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x-6)\left(x^{2}-2 x-12\right)=0 \\
x \geqslant 4
\end{array}\right.\right.\right.
$$

The equation in the system has roots $x=6, x=1 \pm \sqrt{13}$, and of these, the inequality is satisfied by $x=6$ and $x=1+\sqrt{13}$. These are the answers to the problem.