6.093. $\left\{\begin{array}{l}v-u=1, \\ w-v=1, \\ (u-1)^{3}+(v-2)^{3}+(w-3)^{3}=3 .\end{array}\right.$
## Solution.

From the first equation of the system, we find $y=1+u$. Substituting this value of $v$ into the second and third equations of the system, we have

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ w - ( 1 + u ) = 1 , } \\
{ ( u - 1 ) ^ { 3 } + ( 1 + u - 2 ) ^ { 3 } + ( w - 3 ) ^ { 3 } = 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
w-u=2, \\
(u-1)^{3}+(u-1)^{3}+(w-3)^{3}=3
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
u-u=2, \\
2(u-1)^{3}+(w-3)^{3}=3 .
\end{array}\right.
\end{aligned}
$$

From the first equation of the last system, we find $w=2+u$. Substituting this value of $w$ into the second equation of the same system, we get

$$
2(u-1)^{3}+(2+u-3)^{3}=3 \Leftrightarrow(u-1)^{3}=1
$$

from which $u=2$. Then $w=4$, and $v=3$.

Answer: $(2 ; 3 ; 4)$.