Question

IF


$\sqrt{ {x}^{2} + \sqrt[3]{ {x}^{4} {y}^{2} } } \: + \: \sqrt{ {y}^{2} + \sqrt[3]{ {x}^{2} {y}^{4} } } = a$
Then Prove That

${x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} }$
​

Answer 1

$\bold{Given}\longrightarrow \\ \sqrt{ {x}^{2} + \sqrt[3]{ {x}^{4} {y}^{2} } } \: + \sqrt{ {y}^{2} + \sqrt[3]{ {x}^{2} {y}^{4} } } \: = a$

$\bold{To \: prove}\longrightarrow \\ {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} }$

$\bold{Concept \: used}\longrightarrow \\ 1) \sqrt[m]{x} \: = \: {x}^{ \frac{1}{m} }$

$\bold{Proof}\longrightarrow \\ \sqrt{ {x}^{2} + \sqrt[3]{ {x}^{4} {y}^{2} } } + \sqrt{ {y}^{2} + \sqrt[3]{ {x}^{2} {y}^{4} } } \: = \: a$

$= > \sqrt{ {x}^{2} + {( {x}^{4} {y}^{2}) }^{ \frac{1}{3} } } + \sqrt{ {y}^{2} + {( {x}^{2} {y}^{4} ) }^{ \frac{1}{3} } } = a$

$= > \sqrt{ {x}^{2} + {x}^{ \frac{4}{3} } {y}^{ \frac{2}{3} } } \: + \sqrt{ {y}^{2} + {x}^{ \frac{2}{3} } {y}^{ \frac{4}{3} } } = a$

$= > \sqrt{ {x}^{ \frac{4}{3} } ( {x}^{2 - \frac{4}{3} } + {y}^{ \frac{2}{3} } )} \: + \sqrt{ {y}^{ \frac{2}{3} } ( {y}^{2 - \frac{4}{3} } + {x}^{ \frac{2}{3} }) }$

$= > {x}^{ \frac{2}{3} } \sqrt{ {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } } + {y}^{ \frac{2}{3} } \sqrt{ {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } } = a$

$= > \sqrt{ {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } } \: ( {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } ) = a$

$= > {( {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } ) }^{ \frac{1}{2} } \: ( {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } ) ^{1} = a$

$= > {( {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } ) }^{ \frac{3}{2} } = a$

$= > {( {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } ) }^{ \frac{3}{2} \times \frac{2}{3} } = {a}^{ \frac{2}{3} }$

$= > {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} }$

Answer 2

Answer:

Hey mate!!

Step-by-step explanation:

Your required answer is in picture