Assuming that $x,\ y,\ z$ are positive real numbers, simplify the following:$( x^{\frac{-2}{3}}y^{\frac{-1}{2}})^{2})$.

Given: $( x^{\frac{-2}{3}}y^{\frac{-1}{2}})^{2}$, while $x,\ y,\ z$ are positive real numbers.

To do: To simplify  $( x^{\frac{-2}{3}}y^{\frac{-1}{2}})^{2}$.

Solution:

 $( x^{\frac{-2}{3}}.y^{\frac{-1}{2}})^{2}$

$=( x^{( \frac{-2}{3})})^{2}.(y^{( \frac{-1}{2})})^{2}$     [$\because ( a^p.b^q)^m=( a^p)^m.( b^q)^m$]

$=( x^{\frac{-4}{3}}).(y^{\frac{-2}{2}})$        [$\because ( a^m)^n=a^{mn}$]

$=\frac{1}{x^{\frac{4}{3}}.y}$                  [$\because a^{-m}=\frac{1}{a^m}$]

$=\frac{1}{x^{4\times\frac{1}{3}}.y}$

$=\frac{1}{\sqrt[3]{x^{4}}.y}$               [$\because x^{\frac{1}{n}}=\sqrt[n]{x}$]

Thus, $( x^{\frac{-2}{3}}.y^{\frac{-1}{2}})^{2}=\frac{1}{\sqrt[3]{x^{4}}.y}$.

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Updated on: 10-Oct-2022

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