10. (20 points) Let $x, y, z \in \mathbf{R}$ and $x+y+z=0$.

If $\left(x^{2}+y^{2}+z^{2}\right)^{3} \geqslant m x^{2} y^{2} z^{2}$
always holds, find the maximum value of the real number $m$.
10. Let $x=y=1, z=-2$, we get $m \leqslant 54$.

Below we prove that equation (1) always holds when $m=54$. Given $x+y+z=0$, we can assume $x \geqslant 0, y \geqslant 0$. Then
$$
\begin{array}{l}
0 \leqslant 4 x y \leqslant(x+y)^{2}=z^{2} . \\
\text { By } z^{2}=(x+y)^{2} \leqslant 2\left(x^{2}+y^{2}\right) \\
\Rightarrow 3 z^{2} \leqslant 2\left(x^{2}+y^{2}+z^{2}\right) \\
\Rightarrow\left(x^{2}+y^{2}+z^{2}\right)^{3} \geqslant \frac{27}{8} z^{6} \geqslant 54 x^{2} y^{2} z^{2} .
\end{array}
$$

Therefore, the maximum value of $m$ is 54.