6.217. $\left\{\begin{array}{l}x^{4}+x^{2} y^{2}+y^{4}=91 \\ x^{2}+x y+y^{2}=13\end{array}\right.$
## Solution.

Let's rewrite the system of equations as follows:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ ( x ^ { 4 } + y ^ { 4 } ) + x ^ { 2 } y ^ { 2 } = 9 1 , } \\
{ x ^ { 2 } + x y + y ^ { 2 } = 1 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left(x^{2}+y^{2}\right)^{2}-x^{2} y^{2}=91, \\
x^{2}+y^{2}=13-x y
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ ( 1 3 - x y ) ^ { 2 } - x ^ { 2 } y ^ { 2 } = 9 1 , } \\
{ x ^ { 2 } + y ^ { 2 } = 1 3 - x y }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x y = 3 , } \\
{ ( x + y ) ^ { 2 } = 1 3 + x y }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x y=3, \\
(x+y)^{2}=16
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x y=3, \\
x+y= \pm 4 .
\end{array}\right. \\
& (x y=1)
\end{aligned} \Leftrightarrow
$$

This system is equivalent to two systems of equations:

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

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