Question

please help me solve this answer ​

Answer 1

Step-by-step explanation:

very nice question. all the best

Answer 2

Solution :

$\displaystyle \sf{ \bold { To \: Show \: - }} \\ \\ \sf{ \bold { \implies { \dfrac{ {( a^3b - ab^3 ) }^2 + 4a^4 b^4 }{ a^2 b^2 } = { (a^2 + b^2) }^2 }}} \\ \\ \sf{ \bold { \star LHS \: - }} \\ \\ \sf{ \bold { \implies { \dfrac{ {( a^3b - ab^3 ) }^2 + 4a^4 b^4 }{ a^2 b^2 } }}} \\ \\ \sf{ \bold { Simplifying \: the \: numerator \: - }} \\ \\ \sf{ \bold { Numerator \leadsto {( a^3b - ab^3 ) }^2 + 4a^4 b^4 }} \\ \\ \sf{ \bold { \leadsto { [ ab( a^2 - b^2 ) ] }^2 + 4a^4 b^4 }} \\ \\ \sf{ \bold { \leadsto a^2 b^2 { (a^2 - b^2 ) } ^2 + 4a^4b^4 }}$

$\displaystyle \sf{ \bold { Let's \: return \: to \: the \: original \: fraction \: . \: . }} \\ \\ \sf{ \bold { \implies { \dfrac{ a^2 b^2 { (a^2 - b^2 ) } ^2 + 4a^4b^4 }{ a^2 b^2 } }}} \\ \\ \sf{ \bold { Taking \: the \: term \: a^2b^2 \: common \: from \: both \: numerator \: \& \: denominator \: - }}$

$\sf{ \bold { \implies { \dfrac{ a^2 b^2 [ ( a^2 - b^2 ) ^2 + 4a^2 b^2 ] }{ a^2 b^2 } }}} \\ \\ \sf{ \bold { \implies [ ( a^2 - b^2 ) ^2 + 4a^2 b^2 ] }}$

$\sf{ \bold { Expanding \: this \: - }} \\ \\ \sf{ \bold { \implies { a^4 + b^4 - 2a^2 b^2 + 4 a^2 b^2 }}} \\ \\ \sf{ \bold { \implies { a^4 + 2a^2b^2 + b^4 }}} \\ \\ \sf{ \bold { \implies { ( a^2 + b^2 ) ^2 }}} \\ \\ \sf{ \bold { Hence \: Shown }}$