6. Given positive real numbers $x, y, z$ satisfy
$$
x y z + x y + y z + z x + x + y + z = 3 \text{. }
$$

Then the maximum value of $u = x y z (x + y + z)$ is $\qquad$
6. $3(\sqrt[3]{4}-1)^{4}$.

When $x=y=z=\sqrt[3]{4}-1$, $u=3(\sqrt[3]{4}-1)^{4}$.
It is only necessary to prove: $u \leqslant 3(\sqrt[3]{4}-1)^{4}$.
Notice,
$$
\begin{array}{l}
27 u=27 x y z(x+y+z) \\
\leqslant(x+y+z)^{3}(x+y+z)=(x+y+z)^{4} .
\end{array}
$$

Thus, $x+y+z \geqslant \sqrt[4]{27 u}$.
Also, $x y z+\frac{(\sqrt[3]{4}-1)^{2}}{3}(x+y+z) \geqslant 2 \sqrt{\frac{(\sqrt[3]{4}-1)^{2}}{3}} u$,
$$
x y+y z+z x \geqslant \sqrt{3 x y z(x+y+z)}=\sqrt{3 u} \text {, }
$$

Then, from equations (1), (2), and (3),
$$
\begin{array}{l}
3=x y z+x y+y z+z x+x+y+z \\
\geqslant\left[1-\frac{(\sqrt[3]{4}-1)^{2}}{3}\right] \sqrt[4]{27 u}+2 \sqrt{\frac{(\sqrt[3]{4}-1)^{2}}{3} u}+\sqrt{3 u} .
\end{array}
$$

Therefore, $u \leqslant 3(\sqrt[3]{4}-1)^{4}$.
Hence, the maximum value of $u=x y z(x+y+z)$ is $3(\sqrt[3]{4}-1)^{4}$.