6.078. $\left\{\begin{array}{l}x^{3}+y^{3}=35 \\ x+y=5\end{array}\right.$
## Solution.

By the formula for the sum of cubes, we get

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ ( x + y ) ( x ^ { 2 } - x y + y ^ { 2 } ) = 3 5 , } \\
{ x + y = 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+y) \\
x+y=5
\end{array}\left((x+y)^{2}-3 x y\right)=35,\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 5 ( 2 5 - 3 x y ) = 3 5 , } \\
{ x + y = 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 2 5 - 3 x y = 7 , } \\
{ x + y = 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
x y=6, \\
x+y=5 .
\end{array}\right.\right.\right.
\end{aligned}
$$

By Vieta's theorem, the possible solutions are; $\left\{\begin{array}{l}x_{1}=2, \\ y_{1}=3\end{array}\right.$ or $\left\{\begin{array}{l}x_{2}=3, \\ y_{2}=2 .\end{array}\right.$

Answer: $(2 ; 3)(3 ; 2)$.