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

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8. 0 .
$$
\begin{array}{l}
\text { Let } 4^{a}=6^{b}=9^{c}=k \text {. } \\
\text { Then } a=\log _{k} 4, b=\log _{k} 6, c=\log _{k} 9 \\
\Rightarrow \frac{1}{a}-\frac{2}{b}+\frac{1}{c}=\log _{k} 4-2 \log _{k} 6+\log _{k} 9 \\
\quad=\log _{k} 1=0 .
\end{array}
$$