$$
\begin{array}{l}
\text { 9. Let }\left(1+x+x^{2}\right)^{n} \\
=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n}\left(n \in \mathbf{N}_{+}\right) \text {. }
\end{array}
$$

Then the value of $a_{0}+a_{3}+a_{6}+\cdots+a_{3\left[\frac{2 n}{3}\right]}$ is (where, $[x]$ is the greatest integer function of $x$).
$9.3^{n-1}$.
Let $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}$. Then
$$
\begin{array}{l}
\omega^{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}, \\
\omega^{3}=1, \\
\cdots \cdots \\
\omega^{3 k}=1, \\
\omega^{3 k+1}=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}, \\
\omega^{3 k+2}=-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i} . \\
\text { Let } A=a_{0}+a_{3}+a_{6}+\cdots+a_{3\left[\frac{2 n}{3}\right]}, \\
B=a_{1}+a_{4}+a_{7}+\cdots+a_{2 n-1}, \\
C=a_{2}+a_{5}+a_{8}+\cdots+a_{2 n} .
\end{array}
$$

Substituting $1, \omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}, \omega^{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}$ into the given equation, we get
$$
\left\{\begin{array}{l}
A+B+C=3^{n}, \\
A+\omega B+\omega^{2} C=0, \\
A+\omega^{2} B+\omega C=0 .
\end{array}\right.
$$

Adding the three equations, and noting that $\omega+\omega^{2}=-1$, we get
$$
A=3^{n-1} \text {. }
$$