6.220. $\left\{\begin{array}{l}\frac{x^{3}}{y}+x y=40 \\ \frac{y^{3}}{x}+x y=10 .\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0\end{array}\right.$

Multiplying the first equation of the system by $y \neq 0$, and the second by $x \neq 0$, we get $\left\{\begin{array}{l}x^{3}+x y^{2}=40 y, \\ y^{3}+x^{2} y=10 x\end{array} \Leftrightarrow\left\{\begin{array}{l}x\left(x^{2}+y^{2}\right)=40 y, \\ y\left(y^{2}+x^{2}\right)=10 x .\end{array}\right.\right.$

Dividing the first equation of the system by the second, we get $\frac{x}{y}=\frac{4 y}{x}, x^{2}=4 y^{2}$, from which $x_{1}=-2 y, x_{2}=2 y$.

1) $\frac{y^{3}}{-2 y}-2 y \cdot y=10, y^{2}=-4$, no solutions.
2) $\frac{y^{3}}{2 y}+2 y \cdot y=10, y^{2}=4$, from which $y_{1,2}= \pm 2$. Then $x_{1,2}= \pm 4$.

Answer: $(4 ; 2),(-4 ;-2)$.