Prove that:$ \left(\frac{64}{125}\right)^{\frac{-2}{3}}+\frac{1}{\left(\frac{256}{625}\right)^{\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^{0}=\frac{61}{16} $

Given: 

\( \left(\frac{64}{125}\right)^{\frac{-2}{3}}+\frac{1}{\left(\frac{256}{625}\right)^{\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^{0}=\frac{61}{16} \)

To do: 

We have to prove that \( \left(\frac{64}{125}\right)^{\frac{-2}{3}}+\frac{1}{\left(\frac{256}{625}\right)^{\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^{0}=\frac{61}{16} \).

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

LHS $=(\frac{64}{125})^{\frac{-2}{3}}+\frac{1}{(\frac{256}{625})^{\frac{1}{4}}}+(\frac{\sqrt{25}}{\sqrt[3]{64}})^0$

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

$=\frac{4^{3 \times(\frac{-2}{3})}}{5^{3} \times \frac{-2}{3}}+\frac{1}{\frac{4^{4 \times \frac{1}{4}}}{5^{4 \times \frac{1}{4}}}}+1$

$=\frac{4^{-2}}{5^{-2}}+\frac{1}{\frac{4}{5}}+1$

$=\frac{5^2}{4^2}+\frac{5}{4}+1$

$=\frac{25}{16}+\frac{5}{4}+1$

$=\frac{25+20}{16}+1$

$=\frac{45}{16}+1$

$=\frac{45+16}{16}$

$=\frac{61}{16}$

$=$ RHS

Hence proved.     

Tutorialspoint

Updated on: 10-Oct-2022

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