if $ x=\left(\frac{5}{8}\right)^{-2} \times\left(\frac{12}{15}\right)^{-2},$ then the walue of $ x^{-3} $ is

Given:   $ x=\left(\frac{5}{8}\right)^{-2} \times\left(\frac{12}{15}\right)^{-2}$

To do: Find the value of  $ x^{-3} $

Solution:   $x=(\frac{5}{2})^{-2} \times(\frac{12}{15})^{-2}$

$x=(\frac{2}{5})^2\times(\frac{15}{12})^2$

$x=\frac{4}{25}\times\frac{225}{144}$

$x=\frac{36}{9}$

$x=4$  

  Now, $x^{-3} = 4^{-3}$

=$ (\frac{1}{4})^3$

= $\frac{1}{ 64}$

Therefore, The value of $ x^{-3} $ is $\frac{1}{ 64}$

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Updated on: 10-Oct-2022

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