Question

Find the value of :-
$\boldsymbol{ \longrightarrow\bigg\{ \sqrt[3]{{x}^{4}y} × \dfrac{1}{\sqrt[4]{{x}^{2}{y}^{8}}} \bigg\}^{-6} \times \dfrac{x}{y} }$

$\red \odot \: {x}^{-11} . {y}^{-14}$

$\green \odot \: {x}^{ - 10} . {y}^{ - 13}$

$\pink \odot \: {x}^{ - 4} . {y}^{9}$

$\blue \odot \: {x}^{ - 12} . {y}^{ - 15}$​

Answer 1

Answer:

Solution:-

$\boldsymbol{ \longrightarrow\bigg\{ \sf \: \sqrt[3]{{x}^{4}y} × \dfrac{1}{\sqrt[4]{{x}^{2}{y}^{8}}} \bigg\}^{-6} \times \dfrac{x}{y} }$

$\boldsymbol{ \longrightarrow = \bigg\{ {\bigg( \sf{x}^{4}y \bigg) }^{\dfrac{1}{3} } × \dfrac{1}{ {\bigg( \sf{x}^{2} {y}^{8} \bigg) }^{\dfrac{1}{4}}}\bigg\}^{-6} \times \dfrac{x}{y} }$

$\boldsymbol{ \longrightarrow = \bigg\{{ \sf \: {x}^{ \frac{4}{3}} \times {y}^{ \frac{1}{3}} \times \dfrac{1}{ {x}^{ \frac{1}{2}} \times {y}^{2}}}\bigg\}^{-6} \times \dfrac{x}{y} } \\ \\ \boldsymbol{ \longrightarrow = \bigg\{{ \sf \: {x}^{ \frac{4}{3} - \frac{1}{2} } \times {y}^{ \frac{1}{3} - 2}}\bigg\}^{-6} \times \dfrac{x}{y} } \\ \\ \boldsymbol{ \longrightarrow = \bigg\{{ \sf{x}^{ \frac{8 - 3}{6}} \times {y}^{ \frac{1 - 6}{3}}}\bigg\}^{-6} \times \sf \: \dfrac{x}{y} }$

$\boldsymbol{ \longrightarrow = \bigg\{{ \sf \: {x}^{ \frac{5}{6}} \times {y}^{ \sf \: \frac{- 5}{3}}}\bigg\}^{-6} \times \sf \dfrac{x}{y} } \\ \\ \boldsymbol{ \longrightarrow = \bigg\{{ \sf {x}^{ \frac{5}{6} \times ( - 6)} \times {y}^{ \frac{- 5}{3} \times ( - 6)}}\bigg\} \times \dfrac{x}{y} } \\ \\ \boldsymbol{ \longrightarrow = \bigg\{{ \sf{x}^{ - 5} \times {y}^{10}}\bigg\} \times \dfrac{x}{y} } \\ \\ \boldsymbol{ \longrightarrow = \bigg\{{ \sf{x}^{ - 5 + 1} \times {y}^{10 - 1}}\bigg\}}$

$\boldsymbol{ \longrightarrow = \bigg\{{ \sf \red{{x}^{ - 4} \times {y}^{9}}\bigg\}}}$

Hence,

$\sf \: Option \bold{(c)} \: is \: correct.\boldsymbol{ = \bigg\{{ \sf \red{{x}^{ - 4} \times {y}^{9}}\bigg\}}}$

Answer 2

Given :

• ${\sf{\bigg\{ \sqrt[3]{{x}^{4}y} × \dfrac{1}{\sqrt[4]{{x}^{2}{y}^{8}}} \bigg\}^{-6} \times \dfrac{x}{y} }}$

Identities Used :

• Laws of exponents ;

• ${\underline{\boxed{\sf{\color{blue}{a^m\times{a^n}\:=\:a^{m\:+\:n}\:}}}}}$

• ${\underline{\boxed{\sf{\color{purple}{\dfrac{a^m}{a^n}\:=\:a^{m\:-\:n}\:}}}}}$

• ${\underline{\boxed{\sf{\color{orange}{\dfrac{1}{x^n}\:=\:x^{-n}\:}}}}}$

• ${\underline{\boxed{\sf{\color{peru}{(a^m)^n\:=\:a^{m\times{n}}\:}}}}}$

• ${\underline{\boxed{\bf{\color{red}{ {x}^{0} = 1}}}}}$

• ${\underline{\boxed{\sf{\green{ \sqrt[n]{x} = {\bigg(x\bigg) }^{\dfrac{1}{n} } }}}}}$

Solution :

${\sf{ :\implies\bigg\{ \sqrt[3]{{x}^{4}y} × \dfrac{1}{\sqrt[4]{{x}^{2}{y}^{8}}} \bigg\}^{-6} \times \dfrac{x}{y} }}$

${\sf{ :\implies\bigg\{\bigg( {x}^{4}y \bigg)^{\frac{1}{3} } × \dfrac{1}{( {x}^{2} {y}^{8 })^{\frac{1}{4}}}\bigg\}^{-6}~ \times ~\dfrac{x}{y}}}$

${\sf{ :\implies{\bigg\{ {x} ^{\frac{4}{3}} \times {y}^{\frac{1}{3}} \times \dfrac{1}{ {x}^{ \frac{1}{2} ~\times ~{y}^{2}}}\bigg\}^{-6} ~\times~ {\dfrac{x}{y}}}}}$

${\sf{ :\implies \bigg\{ {x}^{ \frac{4}{3} - \frac{1}{2}}~ \times~ {y}^{ \frac{1}{3} - 2}\bigg\}^{-6} ~\times ~\dfrac{x}{y} }}$

${\sf{ :\implies\bigg\{ {x}^{ \frac{8~ - ~3}{6}} ~\times ~{y}^{ \frac{1~ - ~6}{3}}\bigg\}^{-6}~ \times~ \dfrac{x}{y} }}$

${\sf{ :\implies\bigg\{ {x}^{ \frac{5}{6}} ~\times~ {y}^{ \frac{- 5}{3}}\bigg\}^{-6} ~\times~ \dfrac{x}{y} }}$

${\sf{ :\implies\bigg\{ {x}^{ \frac{5}{6}}~ \times ~( - 6) ~\times~ {y}^{ \frac{- 5}{3}}~ \times~ ( - 6)\bigg\} ~\times~ \dfrac{x}{y} }}$

${\sf{ :\implies\bigg\{ {x}^{ - 5} ~\times~ {y}^{10}\bigg\} ~\times~ \dfrac{x}{y} }}$

${\sf{ :\implies\bigg\{ {x}^{ - 5 ~+~ 1} ~\times~ {y}^{10~ - ~1}}\bigg\}}$

• $\boxed{\sf{\frak{\pink{\bigg\{ {x}^{ - 4}~ \times ~{y}^{9}}\pink{\bigg\}}}}}$

$\begin{gathered} \\ \end{gathered}$

Hence,

• The value is ${\sf{\bigg\{{ {x}^{ - 4} \times {y}^{9}}\bigg\}}}$

$\begin{gathered} \\ \rule{190pt}{2pt} \\ \end{gathered}$

Additional Information

$\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$