8. If $4^{a}=6^{b}=9^{c}$, then
$$
\frac{1}{a}-\frac{2}{b}+\frac{1}{c}=
$$
$\qquad$
8. 0 .

Let $4^{a}=6^{b}=9^{c}=k$.
Then $a=\log _{k} 4, b=\log _{k} 6, c=\log _{k} 9$
$$
\begin{array}{l}
\Rightarrow \frac{1}{a}-\frac{2}{b}+\frac{1}{c}=\log _{k} 4-2 \log _{k} 6+\log _{k} 9 \\
\quad=\log _{k} 1=0 .
\end{array}
$$