6. Calculate:
$$
\sum_{x=1}^{9} C_{9}^{x}\left[\sum_{y=0}^{9-x} C_{9-x}^{y}\left(\sum_{z=0}^{9-x-y} C_{9-x-y}^{z}\right)\right]=
$$
$\qquad$
6. $4^{9}-3^{9}$.

Consider the polynomial
$$
\begin{array}{l}
(a+b+c+d)^{9} \\
=\sum_{x=0}^{9}\left[\sum_{y=0}^{9-x}\left(\sum_{z=0}^{9-x-y} \mathrm{C}_{9}^{x} \mathrm{C}_{9-x}^{y} \mathrm{C}_{9-x-y}^{z} a^{x} b^{y} c^{z} d^{9-x-y-z}\right)\right] .
\end{array}
$$

Let $a=b=c=d=1$. Then
$$
\sum_{x=0}^{9} \mathrm{C}_{9}^{x}\left[\sum_{y=0}^{9-x} \mathrm{C}_{9-x}^{y}\left(\sum_{z=0}^{9-x-y} \mathrm{C}_{9-x-y}^{z}\right)\right]=4^{9} \text {. }
$$

Similarly, when $x=0$,
$$
\sum_{y=0}^{g} \mathrm{C}_{9}^{y}\left(\sum_{i=0}^{9-\gamma} \mathrm{C}_{9-y}^{z}\right)=3^{9} \text {. }
$$

Therefore, the original expression $=4^{9}-3^{9}$.