6. Given that for any $x, y, z \geqslant 0$ there is $x^{3}+y^{3}+z^{3}-3 x y z \geqslant c|(x-y)(y-z)(z-x)|$. Find the maximum value of $c$.


6. Left side of the inequality
$$
\begin{array}{l}
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right) \\
=\frac{(x+y+z)\left((x-z)^{2}+(y-z)^{2}+(z-x)^{2}\right)}{2} .
\end{array}
$$

Assume without loss of generality that $x \geqslant y \geqslant z \geqslant 0$.
If $x=y$ or $y=z$, the inequality holds for any real number $c$.

If $x>y>z \geqslant 0$, fix $x-y$ and $y-z$, then $z-x$ is also fixed.

To maximize $c$, the left side should be minimized, i.e., $x+y+z$ should be minimized.

Since $x+y+z=(x-y)+2(y-z)+3z$, the left side is minimized when $z=0$.
Now consider the case $x>y>z=0$.
The original expression becomes $x^{3}+y^{3} \geqslant c x y(x-y)$.
Let $t=\frac{x}{y}(t>1)$. Then $t^{3}+1 \geqslant c t(t-1)$. Hence, $c_{\max }=f(t)_{\min }$, where $f(t)=\frac{t^{3}+1}{t(t-1)}$. Suppose $f(t)$ attains its minimum at $t_{0}$. Then
$$
f^{\prime}\left(t_{0}\right)=\frac{t_{0}^{4}-2 t_{0}^{3}-2 t_{0}+1}{t_{0}^{2}\left(t_{0}-1\right)^{2}}=0 \text {. }
$$

Thus, $\left(t_{0}+\frac{1}{t_{0}}\right)^{2}-2\left(t_{0}+\frac{1}{t_{0}}\right)-2=0$.
Solving this, we get $t_{0}=\frac{1+\sqrt{3}+\sqrt{2 \sqrt{3}}}{2}$.
Therefore, $c_{\max }=f\left(t_{0}\right)=\left(\frac{\sqrt{6}+3 \sqrt{2}}{2}\right) \sqrt[4]{3}$.