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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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