Question

Prove that:

$\\ \\ \bf\bigg [ \bigg( \frac{1}{2} {\bigg)}^{2} {\bigg]}^{3} \times \bigg( \frac{1}{3} {\bigg)}^{ - 4} \times {3}^{ - 2} \times \frac{1}{6} = \frac{3}{128}$
​

Answer 1

Step-by-step explanation:

hope this will help you

Answer 2

Given:

$\large\underline{\boxed{\tt{ { ({( \frac{1}{2} )}^{2} })^{3} \times {( \frac{1}{3}) }^{ - 4} \times {3}^{ - 2} \times \frac{1}{6}=\frac{3}{128} }}}$

To prove:

L.H.S = R.H.S

Solution:

$\large{\sf{L.H.S,}}$

$\large{\sf{ { ({( \frac{1}{2} )}^{2} })^{3} \times {( \frac{1}{3}) }^{ - 4} \times {3}^{ - 2} \times \frac{1}{6} }}$

By using, $\large{\bold{ {(( {a})^{b} )}^{c} = {(a)}^{b \times c} }}$

$\Large\implies{\sf{ {( \frac{1}{2}) }^{6} \times {( \frac{1}{3}) }^{ - 4} \times {3}^{ - 2} \times \frac{1}{6}}}$

By using, $\large{\bold{ { (\frac{a}{b} )}^{ - c} = {( \frac{b}{a} )}^{c} }}$

$\Large\implies{\sf{{( \frac{1}{2}) }^{6} \times {( \frac{3}{1} )}^{4} \times {3}^{ - 2} \times \frac{1}{6}}}$

By using, $\large{\bold{ {(a)}^{ - b} = {( \frac{1}{a}) }^{b} }}$

$\Large\implies{\sf{{( \frac{1}{2}) }^{6} \times {( \frac{3}{1} )}^{4} \times {( \frac{1}{3} )}^{2} \times \frac{1}{6}}}$

By using, $\large{\bold{ { (\frac{a}{b} )}^{c} = ( \frac{ {a}^{c} }{{b}^{c} } )}}$

$\Large\implies{\sf{ \frac{ {1}^{6} }{ {2}^{6} } \times \frac{ {3}^{4} }{ {1}^{4} } \times \frac{ {1}^{2} }{ {3}^{2} } \times \frac{1}{6} }}$

$\Large\implies{\sf{ \frac{1}{64} \times \frac{81}{1} \times \frac{1}{9} \times \frac{1}{6} }}$

$\Large\implies{\sf{ \frac{1}{64} \times \frac{9}{1} \times \frac{1}{1} \times \frac{1}{6} }}$

$\Large\implies{\sf{ \frac{1}{64} \times \frac{3}{1} \times \frac{1}{2} }}$

$\Large\implies{\sf{ \frac{3}{64 \times 2} }}$

$\Large\implies{\boxed{\sf{\red{\frac{3}{128}=R.H.S}}}}$

$\Large{\underline{\underline{\bold{\blue{Hence,\:proved!}}}}}$

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