6.230. $\left\{\begin{array}{l}\sqrt[3]{x+2 y}+\sqrt[3]{x-y+2}=3, \\ 2 x+y=7\end{array}\right.$
Solution.

Raising the first equation of the system to the third power, we get

$$
\begin{gathered}
x+2 y+3 \sqrt[3]{(x+2 y)^{2}(x-y+2)}+3 \sqrt[3]{(x+2 y)(x-y+2)^{2}}+x-y+ \\
+2=27 \Leftrightarrow 2 x+7+3 \sqrt[3]{(x+2 y)(x-y+2)}(\sqrt[3]{x+2 y}+3 \sqrt[3]{x-y+2})=25
\end{gathered}
$$

Using the equations of the system, we obtain

$$
\begin{aligned}
& 7+9 \sqrt[3]{(x+2 y)(x-y+2)}=25 \Leftrightarrow \sqrt[3]{(x+2 y)(x-y+2)}=2 \Leftrightarrow \\
& \Leftrightarrow(x+2 y)(x-y+2)=8 . \text { We have }\left\{\begin{array}{l}
(x+2 y)(x-y+2)=8, \\
2 x+y=7
\end{array} \Rightarrow\right. \\
& y=7-2 x,(x+2(7-2 x))(x-7+2 x+2)=8,9 x^{2}-57 x+78=0,
\end{aligned}
$$

from which $x_{1}=2, x_{2}=\frac{13}{3}$. Then $y_{1}=3, y_{2}=-\frac{5}{3}$.

Answer: $(2 ; 3),\left(\frac{13}{3} ;-\frac{5}{3}\right)$.