Example 3 If $\left(1+x+x^{2}+x^{3}\right)^{5}\left(1-x+x^{2}-\right.$ $\left.x^{3}\right)^{5}=a_{30}+a_{29} x+\cdots+a_{1} x^{29}+a_{0} x^{30}$, find $a_{15}$.
Let $f(x)=\left(1+x+x^{2}+x^{3}\right)^{5}$, then the original expression is
$$
F(x)=f(x) f(-x)
$$

which is an even function. Therefore, we have
$$
\begin{array}{l}
\boldsymbol{F}(x)=\frac{1}{2}[\boldsymbol{F}(x)+\boldsymbol{F}(-x)] \\
=a_{30}+a_{28} x^{2}+\cdots+a_{2} x^{28}+a_{0} x^{30},
\end{array}
$$

where all the coefficients of the odd powers of $x$ are zero, hence $a_{15}=0$.