Question

solve this easy Question fastly solve
${a}^{4} + {b}^{4} = ( {a}^{2} + pab + {b}^{2} )( {a}^{2} - qab + {b}^{2} )$


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Answer 1

Step-by-step explanation:

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

Q:-solve this and then find pq

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

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

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

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

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

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

On comparing both sides :-

we get p=2 & q =2

$∴pq = 2 {x}^{2} = 4$

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HOPE IT HELPS YOU..

_____________________

Thankyou:)

Answer 2

\

Q:-solve this and then find pq

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

4

+b

4

=(a

2

+pab+b

2

)(a

2

−qab+b

2

)

\huge\tt\underline\blue{Answer }

Answer

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

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

4

+b

4

=(a

2

)

2

+(b

2

)

2

⟹ {( {a}^{2} + {b}^{2} ) }^{2} - 4 {a}^{2} {b}^{2} [tex]⟹ { ({a}^{2} + {b}^{2} )}^{2} - ( {2ab)}^{2}⟹(a

2

+b

2

)

2

−4a

2

b

2

[tex]⟹(a

2

+b

2

)

2

−(2ab)

2

⟹ {a}^{4} + {b}^{4} = ( {a}^{2} + {b}^{2} + 2ab)( {a}^{2} + {b}^{2} - 2ab)⟹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} )⟹(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∴pq=2x

2

=4

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HOPE IT HELPS YOU..

_____________________

Thankyou:)