6. Let $0 \leqslant x_{i} \leqslant 1(i=1,2, \cdots, 5)$. Then
$$
\begin{array}{l}
\left|x_{1}-x_{2}\right|^{3}+\left|x_{2}-x_{3}\right|^{3}+\left|x_{3}-x_{4}\right|^{3}+ \\
\left|x_{4}-x_{5}\right|^{3}+\left|x_{5}-x_{1}\right|^{3}
\end{array}
$$

The maximum value is . $\qquad$
6. 4 .

It is easy to know that there exist three consecutive numbers, which are either increasing or decreasing in sequence.
Assume $x_{1} \leqslant x_{2} \leqslant x_{3}$. System $\mathrm{Q}$ : 36022
$$
\begin{array}{l}
\left|x_{1}-x_{2}\right|^{3}+\left|x_{2}^{2}-x_{3}\right|^{3} \\
\leqslant\left(x_{2}-x_{1}\right)+\left(x_{3}-x_{2}\right)=x_{3}-x_{1} \leqslant 1 \text {. } \\
\end{array}
$$

And $\left|x_{3}-x_{4}\right|^{3} 、\left|x_{4}-x_{5}\right|^{3} 、\left|x_{5}-x_{1}\right|^{3} \leqslant 1$, so
$$
\begin{array}{l}
\left|x_{1}-x_{2}\right|^{3}+\left|x_{2}-x_{3}\right|^{3}+\left|x_{3}-x_{4}\right|^{3}+ \\
\left|x_{4}-x_{5}\right|^{3}+\left|x_{5}-x_{1}\right|^{3} \leqslant 4 .
\end{array}
$$

When $x_{1}=x_{3}=x_{5}=1, x_{2}=x_{4}=0$,
$$
\begin{array}{l}
\left|x_{1}-x_{2}\right|^{3}+\left|x_{2}-x_{3}\right|^{3}+\left|x_{3}-x_{4}\right|^{3}+ \\
\left|x_{4}-x_{5}\right|^{3}+\left|x_{5}-x_{1}\right|^{3}=4 .
\end{array}
$$

Therefore, the maximum value sought is 4 .