6.227. $\left\{\begin{array}{l}\sqrt[3]{u+v}+\sqrt[3]{v+w}=3 \\ \sqrt[3]{v+w}+\sqrt[3]{w+u}=1 \\ \sqrt[3]{w+u}+\sqrt[3]{u+v}=0\end{array}\right.$
## Solution.

$$
\text { Let }\left\{\begin{array}{l}
\sqrt[3]{u+v}=x \\
\sqrt[3]{v+w}=y \\
\sqrt[3]{w+u}=z
\end{array}\right.
$$

With respect to $x, y, z$, the system takes the form $\left\{\begin{array}{l}x+y=3, \\ y+z=1, \\ z+x=0\end{array} \Leftrightarrow\left\{\begin{array}{l}x+y=3, \\ y-x=1, \\ z=-x .\end{array}\right.\right.$

By adding and subtracting the first and second equations, we get $x=1 ; y=2$.

$$
\begin{gathered}
\text { Then } z=-1 . \text { Hence }\left\{\begin{array} { l } 
{ \sqrt [ 3 ] { u + v } = 1 , } \\
{ \sqrt [ 3 ] { v + w } = 2 , } \\
{ \sqrt [ 3 ] { w + u } = - 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
u+v=1, \\
v+w=8, \\
w+u=-1
\end{array}, \Leftrightarrow\left\{\begin{array}{l}
u+v=1, \\
v+w=8, \\
w=-1-u
\end{array} \Leftrightarrow\right.\right.\right. \\
\Leftrightarrow\left\{\begin{array}{l}
u+v=1, \\
v-1-u=8 \\
w=-1-u
\end{array}, \Leftrightarrow\left\{\begin{array}{l}
u+v=1, \\
v-u=9, \\
w=-1-u,
\end{array}, u=-4 ; v=5 ; w=3 .\right.\right.
\end{gathered}
$$

Answer: $(-4 ; 5 ; 3)$.