3. Solve the system of equations

$$
\left\{\begin{array}{l}
x^{3}+y^{3}=341 \\
x y+y x^{2}=330
\end{array}\right.
$$
Solution. If we multiply the second equation by 3, and then add the two equations, we get $(x+y)^{3}=11^{3}$, from which $x+y=11$. Substituting into the second equation, we get $xy=30$. In the end, we have $x=5, y=6$ or $x=6, y=5$.