If $ \left(\frac{a}{b}\right)=\left(\frac{5}{2}\right)^{-3} \times\left(\frac{8}{15}\right)^{-3} $ then $ \left(\frac{a}{b}\right)^{-2} $ is equal to


Given:

\( \left(\frac{a}{b}\right)=\left(\frac{5}{2}\right)^{-3} \times\left(\frac{8}{15}\right)^{-3} \)

To do:

We have to find \( \left(\frac{a}{b}\right)^{-2} \).
Solution:
 We know that,

$a^m \times b^m=(a\times b)^m$

$\frac{a}{b}=(\frac{5}{2})^{-3} \times(\frac{8}{15})^{-3}$

$=(\frac{5}{2}\times\frac{8}{15})^{-3}$

$=(\frac{4}{3})^{-3}$

$(\frac{a}{b})^{-2}=[(\frac{4}{3})^{-3}]^{-2}$

$=(\frac{4}{3})^{-3\times-2}$

$=(\frac{4}{3})^{6}$

Therefore,

$(\frac{a}{b})^{-2}=(\frac{4}{3})^{6}$.

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Updated on: 10-Oct-2022

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