6.071. $\left\{\begin{array}{l}x-y=1, \\ x^{3}-y^{3}=7 .\end{array}\right.$
## Solution.

Transform the second equation of the system

$\left\{\begin{array}{l}x-y=1, \\ (x-y)\left(x^{2}+x y+y^{2}\right)=7\end{array} \Leftrightarrow\left\{\begin{array}{l}x-y=1, \\ (x-y)^{2}+3 x y=7\end{array}\right.\right.$
$\left\{\begin{array}{l}x-y=1, \\ 1+3 x y=7\end{array} \Leftrightarrow\left\{\begin{array}{l}x-y=1, \\ x y=2,\end{array}\right.\right.$ 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: $(2 ; 1)(-1 ;-2)$.