Find the value of $( \frac{125}{64})^2+( \frac{1}{( \frac{625}{256})^{-\frac{1}{4}}})+[( \frac{\sqrt{36}}{\sqrt[3]{64}})^0]^{\frac{1}{2}}$.

Given: $( \frac{125}{64})^2+( \frac{1}{( \frac{625}{256})^{-\frac{1}{4}}})+[( \frac{\sqrt{36}}{\sqrt[3]{64}})^0]^{\frac{1}{2}}$

To do: To find the value $( \frac{125}{64})^2+( \frac{1}{( \frac{625}{256})^{-\frac{1}{4}}})+[( \frac{\sqrt{36}}{\sqrt[3]{64}})^0]^{\frac{1}{2}}$.

Solution: 

$( \frac{125}{64})^2+( \frac{1}{( \frac{625}{256})^{-\frac{1}{4}}})+[( \frac{\sqrt{36}}{\sqrt[3]{64}})^0]^{\frac{1}{2}}$

$=( \frac{5^3}{4^3})^2+( \frac{5^4}{4^4})^{\frac{1}{4}}+[1]^{\frac{1}{2}}$

$=( \frac{5^3}{4^3})^2+( \frac{(5^4)^{\frac{1}{4}}}{( 4^4)^{\frac{1}{4}}}+1$         [$\because 1^x=1$]

$=\frac{( 5)^{3\times2}}{( 4)^{3\times2}}+\frac{( 5)^{( 4\times\frac{1}{4})}}{( 4)^{( 4\times\frac{1}{4})}}+1$

$=\frac{5^6}{4^6}+\frac{5}{4}+1$

$=\frac{15,625}{4,096}+\frac{5}{4}+1$

$=3.80+1.25+1$

$=6.05$

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Updated on: 10-Oct-2022

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