6.115. $\left\{\begin{array}{l}\sqrt[3]{x}+\sqrt[3]{y}=3 \\ \sqrt[3]{x^{2}}-\sqrt[3]{x y}+\sqrt[3]{y^{2}}=3\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
\sqrt[3]{x} + \sqrt[3]{y} = 3 \\ 
\sqrt[3]{x^{2}} - \sqrt[3]{x y} + \sqrt[3]{y^{2}} = 3
\end{array}\right.
\]
Solution.

Let $\left\{\begin{array}{l}\sqrt[3]{x}=u, \\ \sqrt[3]{y}=v,\end{array}\left\{\begin{array}{l}x=u^{3}, \\ y=v^{3}\end{array}\right.\right.$

With respect to $u$ and $v$, the system takes the form

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ u + v = 3 , } \\
{ u ^ { 2 } - u v + v ^ { 2 } = 3 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ u + v = 3 , } \\
{ ( u + v ) ^ { 2 } - 3 u v = 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
u+v=3 \\
3^{2}-3 u v=3
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
u+v=3 \\
u v=2
\end{array}\right.
\end{aligned}
$$

from which $\left\{\begin{array}{l}u_{1}=1, \\ v_{1}=2 ;\end{array}\left\{\begin{array}{l}u_{2}=2, \\ v_{2}=1 .\end{array}\right.\right.$ Then $\left\{\begin{array}{l}\sqrt[3]{x}=1, \\ \sqrt[3]{y}=2,\end{array}\right.$ from which $\left\{\begin{array}{l}x_{1}=1, \\ y_{1}=8 ;\end{array}\right.$

$$
\left\{\begin{array}{l}
\sqrt[3]{x}=2, \\
\sqrt[3]{y}=1,
\end{array}\right.
$$

Answer: $(1 ; 8),(8 ; 1)$.