Question

Q:-solve this and then find pq
${a}^{4} + {b}^{4} = ( {a}^{2} + pab + {b}^{2} )( {a}^{2} - qab + {b}^{2} )$
​

Answer 1

$\green{\bold{\underline{ ✪ UPSC-ASPIRANT✪ }}}$

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-solve this and then find pq

${a}^{4} + {b}^{4} = ( {a}^{2} + pab + {b}^{2} )( {a}^{2} - qab + {b}^{2} )$

$\huge\tt\underline\blue{Answer }$

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

══════════XXX═════════════

$⟹ {a}^{4} + {b}^{4} = { ({a}^{2} )}^{2} + {( {b}^{2} )}^{2}$

$⟹ {( {a}^{2} + {b}^{2} ) }^{2} - 4 {a}^{2} {b}^{2} </p><p></p><p>[tex]⟹ { ({a}^{2} + {b}^{2} )}^{2} - ( {2ab)}^{2}$

$⟹ {a}^{4} + {b}^{4} = ( {a}^{2} + {b}^{2} + 2ab)( {a}^{2} + {b}^{2} - 2ab)$

$⟹( {a}^{2} + 2ab + {b}^{2} )( {a}^{2} - 2ab + {b}^{2} ) = ( {a}^{2} + pab + {b}^{2} )( {a}^{2} - qab + {b}^{2} )$

On comparing both sides :-

we get p=2 & q =2

$∴pq = 2 {x}^{2} = 4$

══════════XXX═════════════

HOPE IT HELPS YOU..

_____________________

Thankyou:)