6.209. $\left\{\begin{array}{l}x^{3}+y^{3}=19 \\ x^{2} y+x y^{2}=-6\end{array}\right.$
## Solution.

From the condition we have $\left\{\begin{array}{l}(x+y)\left((x+y)^{2}-3 x y\right)=19, \\ x y(x+y)=-6 .\end{array}\right.$

Let $\left\{\begin{array}{l}x+y=u, \\ x y=v .\end{array}\right.$ Relative to $u$ and $v$, the system takes the form

$$
\left\{\begin{array}{l}
u\left(u^{2}-3 v\right)=19, \\
u v=-6
\end{array} \Rightarrow u^{3}=1, \text { from which } u=1, v=-6\right.
$$

We obtain $\left\{\begin{array}{l}x+y=1, \\ x y=-6,\end{array} \Rightarrow\left\{\begin{array}{l}x_{1}=-2, \\ y_{1}=3 ;\end{array}\left\{\begin{array}{l}x_{2}=3, \\ y_{2}=-2 .\end{array}\right.\right.\right.$

Answer: $(-2 ; 3),(3 ;-2)$.