factorize a³ +3a²b + 3ab² + b³ - 8
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dekshaverma please ask my ans for correction
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Answered by
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=a³ + 3a² b + 3a b² + b^³ = 8
={a³ + 2a² b + ab²} + { a² b + 2a b² + b²} = 8
=a{a² + 2ab + b²} + b{a² + 2ab + b²} = 8
= {a + b} {a² + 2ab + b²} = 8
= { a + b } { a² + ab + ab + b² } = 8
{ a + b } { a ( a + b ) + b ( a + b ) } = 8
= { ( a + b )( a + b )² } = 8
= ( a + b )³ = 8
∴ a + b = 2
={a³ + 2a² b + ab²} + { a² b + 2a b² + b²} = 8
=a{a² + 2ab + b²} + b{a² + 2ab + b²} = 8
= {a + b} {a² + 2ab + b²} = 8
= { a + b } { a² + ab + ab + b² } = 8
{ a + b } { a ( a + b ) + b ( a + b ) } = 8
= { ( a + b )( a + b )² } = 8
= ( a + b )³ = 8
∴ a + b = 2
Answered by
42
hai friend,
here is your answer
(a+b)³=a³+b³+3ab²+3a²b
thus ,
=a³ +3a²b + 3ab² + b³ - 8
= (a+b)³-8
we know that 8=2³
and a³-b³= a - b)(a2 + ab + b2)
here a=a+b
b=8
thus,
=a³ +3a²b + 3ab² + b³ - 8
=(a+b-2)[(a+b)²+2(a+b)+2²]
= (a+b-2)[(a+b)(a+b+2)+4]
hope helped :)
here is your answer
(a+b)³=a³+b³+3ab²+3a²b
thus ,
=a³ +3a²b + 3ab² + b³ - 8
= (a+b)³-8
we know that 8=2³
and a³-b³= a - b)(a2 + ab + b2)
here a=a+b
b=8
thus,
=a³ +3a²b + 3ab² + b³ - 8
=(a+b-2)[(a+b)²+2(a+b)+2²]
= (a+b-2)[(a+b)(a+b+2)+4]
hope helped :)
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