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\[
\frac{9}{16b^2}
\]
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To simplify \( \left(\frac{4 b}{3}\right)^{-2} \), we can use the property of exponents that states \( a^{-n} = \frac{1}{a^n} \). This means we can rewrite the expression as: \[ \left(\frac{4 b}{3}\right)^{-2} = \frac{1}{\left(\frac{4 b}{3}\right)^{2}} = \frac{1}{\frac{(4 b)^2}{3^2}} = \frac{1}{\frac{16 b^2}{9}} = \frac{9}{16 b^2} \] So, the final answer is: \[ \frac{9}{16 b^2} \]