6.189. $\left\{\begin{array}{l}(x-y)\left(x^{2}+y^{2}\right)=5, \\ (x+y)\left(x^{2}-y^{2}\right)=9 .\end{array}\right.$
## Solution.

Let's rewrite the system of equations as

$$
\left\{\begin{array}{l}
x^{3} + x y^{2} - x^{2} y - y^{3} = 5 \\
x^{3} - x y^{2} + x^{2} y - y^{3} = 9
\end{array}\right.
$$

By adding and subtracting the first and second equations, we get

$\left\{\begin{array}{l}2 x^{3} - 2 y^{3} = 14, \\ 2 x y^{2} - 2 x^{2} y = -4\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{3} - y^{3} = 7, \\ -x y(x - y) = -2\end{array} \Leftrightarrow\left\{\begin{array}{l}(x - y)\left((x - y)^{2} + 3 x y\right) = 7, \\ x y(x - y) = 2 .\end{array}\right.\right.\right.$

Let $\left\{\begin{array}{l}x - y = u, \\ x y = v .\end{array}\right.$ Then the system of equations in terms of $u$ and $v$ becomes

$\left\{\begin{array}{l}u(u + 3 v) = 7, \\ u v = 2 .\end{array} \Rightarrow v = \frac{2}{u}\right.$, hence

$u\left(u^{2} + 3 \cdot \frac{2}{u}\right) = 7 \Leftrightarrow u^{3} = 1, u = 1 ; v = 2$.

Then $\left\{\begin{array}{l}x - y = 1, \\ x y = 2,\end{array}\right.$ from which $\left\{\begin{array}{l}x_{1} = 2, \\ y_{1} = 1 ;\end{array}\left\{\begin{array}{l}x_{2} = -1, \\ y_{2} = -2 .\end{array}\right.\right.$

Answer: $(2 ; 1), (-1 ; -2)$.