6.077. $\left\{\begin{array}{l}x^{4}+y^{4}=82, \\ x y=3\end{array}\right.$
## Solution.

Rewrite the system as

$$
\left\{\begin{array} { l } 
{ ( ( x + y ) ^ { 2 } - 2 x y ) ^ { 2 } - 2 x ^ { 2 } y ^ { 2 } = 82, } \\
{ x y = 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left((x+y)^{2}-6\right)^{2}-18=82, \\
x y=3
\end{array} \Leftrightarrow\right.\right.
$$

$\Leftrightarrow\left\{\begin{array}{l}\left((x+y)^{2}-6\right)^{2}=100 \\ x y=3\end{array}\right.$

From the first equation $\left|(x+y)^{2}-6\right|=10$, hence $(x+y)^{2}=16$ or $(x+y)^{2}=-4$; $(x+y)^{2}=-4$ is not valid. Therefore,

1) $\left\{\begin{array}{l}x+y=4, \\ x y=3,\end{array},\left\{\begin{array}{l}x_{1}=1, \\ y_{1}=3,\end{array},\left\{\begin{array}{l}x_{2}=3 \\ y_{2}=1\end{array}\right.\right.\right.$

or
2) $\left\{\begin{array}{l}x+y=-4, \\ x y=3,\end{array},\left\{\begin{array}{l}x_{3}=-1, \\ y_{3}=-3,\end{array}\left\{\begin{array}{l}x_{4}=-3 \\ y_{4}=-1\end{array}\right.\right.\right.$

Answer: $(1 ; 3),(3 ; 1)(-1,-3)(-3,-1)$.