6.081. $\left\{\begin{array}{l}x^{3}+y^{3}=65 \\ x^{2} y+x y^{2}=20\end{array}\right.$
## Solution.

By expanding the left sides, we can represent the system as

$$
\left\{\begin{array} { l } 
{ ( x + y ) ( x ^ { 2 } - x y + y ^ { 2 } ) = 6 5 , } \\
{ x y ( x + y ) = 2 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+y)\left((x+y)^{2}-3 x y\right)=65 \\
x y(x+y)=20
\end{array}\right.\right.
$$

Let $\left\{\begin{array}{l}x+y=u, \\ x y=v .\end{array}\right.$ Then $\left\{\begin{array}{l}u\left(u^{2}-3 v\right)=65, \\ u v=20\end{array} \Leftrightarrow\left\{\begin{array}{l}u\left(u^{2}-3 v\right)=65, \\ v=\frac{20}{u} .\end{array}\right.\right.$

From the first equation, we get $u\left(u^{2}-3 \cdot \frac{20}{u}\right)=65, u^{3}=125$, from which $u=5$. Then $v=\frac{20}{5}=4$ and $\left\{\begin{array}{l}x+y=5, \\ x y=4 .\end{array}\right.$ From this, $\left\{\begin{array}{l}x_{1}=4, \\ y_{1}=1\end{array}\right.$ and $\left\{\begin{array}{l}x_{2}=1, \\ y_{2}=4 .\end{array}\right.$

Answer: $(4 ; 1)(1 ; 4)$