Question

Simplify and Express the result in positive exponent [∛x⁴y × 1/∛xy⁷]^-4

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Answer 1

GIVEN :-

$\implies\tt \bigg [\sqrt[3]{x^4y} \times \dfrac{1}{\sqrt[3]{xy^7}} \bigg]$

SOLUTION :-

$\tt [\sqrt[3]{x^4y} \times \dfrac{1}{\sqrt[3]{xy^7}} ]^{-4} \\\\\\ \tt Using \ \ identity \ \ \star \sqrt[n]{x^m} = x^{m/n} \\\\\\ \tt \Rightarrow [(x^4y)^{1/3} \times \dfrac{1}{(xy^7)^{1/3} }]^{-4} \\\\\\ \tt By \ \ using \ \ identity \ \ \star (x^my^n)^p = (x)^{mp} \times (y)^{np} \\\\\\ \tt \Rightarrow [\dfrac{(x)^{4/3}\times (y)^{1/3}}{(x)^{2/3}\times (y)^{7/3}}]\\\\\\ \tt \Rightarrow [\dfrac{(x)^{(4/3 - 1/3)}}{(y)^{(7/3 - 1/3)}}]^{-4} \\\\\\ \tt \Rightarrow [\dfrac{(x)^{(\frac{4-1}{3})}}{y^{^\frac{7-1}{3}}}]^{-4}\\\\\\ \tt \Rightarrow [\dfrac{(x)^\frac{3}{3}}{y^{\frac{6}{3}}}]^{-4} \\\\\\ \tt \Rightarrow [\dfrac{(x)^\cancel{\frac{3}{3}}}{(y)^{\cancel{\frac{6}{3}}}}]^{-4} \\\\\\\tt\Rightarrow [\dfrac{x}{y^2}]^{-4} \\\\\\ \tt\Rightarrow [\dfrac{y^2}{x}]^{4}\\\\ \tt \Rightarrow\dfrac{y^8}{x^4}$

Hence Answer is y⁸/x⁴

Answer 2

ANSWER:

$\bigg({\sqrt[3]{x^4y}\times{ \dfrac{1}{ \sqrt[3]{xy^7}}}} \: \bigg)^{ - 4} = \dfrac{ {y}^{8} }{ {x}^{4} }$

GIVEN:

$\bigg({\sqrt[3]{x^4y}\times{ \dfrac{1}{ \sqrt[3]{xy^7}}}} \: \bigg)^{ - 4}$

TO EXPRESS:

$\bigg({\sqrt[3]{x^4y}\times{ \dfrac{1}{ \sqrt[3]{xy^7}}}} \: \bigg)^{ - 4} in \: positive \: exponent$

EXPLANATION:

$\bigg({\sqrt[3]{x^4y}\times{ \dfrac{1}{ \sqrt[3]{xy^7}}}} \: \bigg)^{ - 4}$

$\bigg( \dfrac{\sqrt[3]{x^4y}}{ \sqrt[3]{xy^7}} \: \bigg)^{ - 4}$

$\bigg( \dfrac{\sqrt[3]{xy^7}}{ \sqrt[3]{x^4y}} \: \bigg)^{4}$

$\left(\sqrt[3]{ \dfrac{xy^7}{ x^4y}} \: \right)^{4}$

${\boxed {\boxed{\bold{\large{\dfrac{ {x}^{m} }{{x}^{n} } = {x}^{m - n} }}}}}$

$\left(\sqrt[3]{ \dfrac{y^6}{ x^{3} }} \: \right)^{4}$

$\sqrt[3]{ {y}^{6} } = {y}^{2}$

$\sqrt[3]{ {x}^{3} } = x$

$\bigg( { \dfrac{ {y}^{2} }{x} } \bigg)^{4}$

$\dfrac{ {y}^{8} }{ {x}^{4} }$

${\boxed {\boxed{\bold{\large{\bigg({\sqrt[3]{x^4y} \times{ \dfrac{1}{ \sqrt[3]{xy^7}}}} \: \bigg)^{ - 4} = \dfrac{ {y}^{8} }{ {x}^{4} } }}}}}$