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Evaluate ${{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}} \div {(\frac{1}{4})}^{- 3}$ .

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\frac{19}{64}

\frac{27}{16}

\frac{64}{19}

\frac{16}{25}

answer is A.

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Detailed Solution

We have to find the value of

\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} \div {(\frac{1}{4})}^{- 3} .

We know according to the law of exponents

a^{- m} = (\frac{1}{a^{m}})

, where m is an integer and a being any real number.
Using the above law,

\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} \div {(\frac{1}{4})}^{- 3}

= \{{(\frac{3}{1})}^{3} - {(\frac{2}{1})}^{3}\} \div {(\frac{4}{1})}^{3}

\Rightarrow (3^{3} - 2^{3}) \div 4^{3}

We know that

3^{3}

27, 2^{3}

8

and

4^{3}

64

.
On substituting we get

\Rightarrow (3^{3} - 2^{3}) \div 4^{3} = (27 - 8) \div 64

\Rightarrow 19 \div 64

\Rightarrow \frac{19}{64}

Therefore,

\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} \div {(\frac{1}{4})}^{- 3} = \frac{19}{64}

.
Hence, option 1 is correct.

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