Prove that:$ \left(\frac{1}{4}\right)^{-2}-3 \times 8^{\frac{2}{3}} \times 4^{0}+\left(\frac{9}{16}\right)^{\frac{-1}{2}}=\frac{16}{3} \

Given: 

\( \left(\frac{1}{4}\right)^{-2}-3 \times 8^{\frac{2}{3}} \times 4^{0}+\left(\frac{9}{16}\right)^{\frac{-1}{2}}=\frac{16}{3} \)

To do: 

We have to prove that \( \left(\frac{1}{4}\right)^{-2}-3 \times 8^{\frac{2}{3}} \times 4^{0}+\left(\frac{9}{16}\right)^{\frac{-1}{2}}=\frac{16}{3} \).

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 $=(\frac{1}{4})^{-2}-3 \times 8^{\frac{2}{3}} \times 4^{0}+(\frac{9}{16})^{\frac{-1}{2}}$

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

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

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

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

$=(2)^{4}-12+\frac{4}{3}$

$=16-12+\frac{4}{3}$

$=4+\frac{4}{3}$

$=\frac{16}{3}$

$=$ RHS

Hence proved. 

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Updated on: 10-Oct-2022

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