6.087. $\left\{\begin{array}{l}x^{4}-y^{4}=15 \\ x^{3} y-x y^{3}=6 .\end{array}\right.$
## Solution.

Let $t=\frac{y}{x}$, then $y=t x$ and the system becomes

$\left\{\begin{array}{l}x^{4}-t^{4} x^{4}=15, \\ x^{4} t-x^{4} t^{3}=6\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{4}\left(1-t^{4}\right)=15, \\ x^{4}\left(t-t^{3}\right)=6 .\end{array}\right.\right.$

After dividing, we get $\frac{1-t^{4}}{t-t^{3}}=\frac{15}{6} \Leftrightarrow 2 t^{2}-5 t+2=0$, from which $t_{1}=\frac{1}{2}, t_{2}=2$.

For $t_{1}=\frac{1}{2}$, from the equation $x^{4}\left(1-t^{4}\right)=15$ we have

$$
x^{4}\left(1-\frac{1}{16}\right)=15 \Leftrightarrow x^{4}=16
$$

from which $x_{1}=-2, x_{2}=2$; then $y_{1}=-1, y_{2}=1$.

For $t_{2}=2$ we have $x^{4}(1-16)=15, x^{4}=-1$, this solution is not valid.

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