6.213. $\left\{\begin{array}{l}x^{6}+y^{6}=65, \\ x^{4}-x^{2} y^{2}+y^{4}=13 .\end{array}\right.$
## Solution.

By the formula for the sum of cubes, we get

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ ( x ^ { 2 } + y ^ { 2 } ) ( x ^ { 4 } - x ^ { 2 } y ^ { 2 } + y ^ { 4 } ) = 6 5 , } \\
{ x ^ { 4 } - x ^ { 2 } y ^ { 2 } + y ^ { 4 } = 1 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left(x^{2}+y^{2}\right) 13=65, \\
x^{4}-x^{2} y^{2}+y^{4}=13
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 5 , } \\
{ x ^ { 4 } - x ^ { 2 } y ^ { 2 } + y ^ { 4 } = 1 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{2}=5, \\
\left(x^{2}+y^{2}\right)^{2}-3 x^{2} y^{2}=13
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 5 , } \\
{ 2 5 - 3 x ^ { 2 } y ^ { 2 } = 1 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{2}=5, \\
x^{2} y^{2}=4 .
\end{array}\right.\right.
\end{aligned}
$$

The system is equivalent to the combination of two systems:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ y ^ { 2 } = 5 - x ^ { 2 } , } \\
{ x ^ { 2 } = 1 }
\end{array} , \text { or } \left\{\begin{array}{l}
y^{2}=5-x^{2}, \\
x^{2}=4
\end{array} \Rightarrow\right.\right. \\
& \Rightarrow\left\{\begin{array} { l } 
{ x _ { 1 } = 2 , } \\
{ y _ { 1 } = 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 2 } = - 2 , } \\
{ y _ { 2 } = - 1 ; }
\end{array} \left\{\begin{array}{l}
x_{3}=2 \\
y_{3}=-1
\end{array}\right.\right.\right. \\
& \left\{\begin{array} { l } 
{ x _ { 4 } = - 2 , } \\
{ y _ { 4 } = 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 5 } = 1 , } \\
{ y _ { 5 } = 2 ; }
\end{array} \left\{\begin{array}{l}
x_{6}=-1, \\
y_{6}=-2 ;
\end{array}\right.\right.\right. \\
& \left\{\begin{array} { l } 
{ x _ { 7 } = 1 , } \\
{ y _ { 7 } = - 2 ; }
\end{array} \left\{\begin{array}{l}
x_{8}=-1 \\
y_{8}=2
\end{array}\right.\right.
\end{aligned}
$$

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