Question

Prove the identity (x+y)cube = x cube + y cube +3xy(x+y) Please answer fast its really urgent..

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Answer 1

Given identity to prove

( x + y )³ = x³ + y³ + 3xy( x + y )

Consider LHS

= ( x + y )³

It can be written as

= ( x + y )¹ ⁺ ²

Since a^( m + n ) = a^m × a^n

= ( x + y ) × ( x + y )²

Using algebraic identity ( x + y )² = x² + y² + 2xy

= ( x + y )( x² + y² + 2xy )

= x( x² + y² + 2xy ) + y( x² + y² + 2xy )

= x³ + xy² + 2x²y + x²y + y³ + 2xy²

Adding like terms

= x³ + y³ + 3x²y + 3xy²

Taking 3xy common in the expression ( 3x²y + 3xy² )

= x³ + y³ + 3xy( x + y )

= RHS

Hence proved.

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$: \implies{\sf{ (x + y)^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

${\bf{\blue{\underline{Now:}}}}$

To prove this we know that,

$: \implies{\sf{ (x + y) ^{2} = {x}^{2} + 2xy + {y}^{2} }}$

And,

$: \implies{\sf{ (x + y) ^{3} = (x + y)(x + y)^{2} }}$

$: \implies{\sf{ (x + y) ^{2} = (x + y)( {x}^{2} + 2xy + {y}^{2} ) }}$

$: \implies{\sf{ (x + y) ^{2} = x ( {x}^{2} + 2xy + {y}^{2} ) + y( {x}^{2} + 2xy + {y}^{2} )}}$

$: \implies{\sf{ {x}^{3} + 2 {x}^{2}y + x {y}^{2} + y {x}^{2} + 2x {y}^{2} + {y}^{3} }}$

And,

$: \implies{\sf{ (x + y) ^{3} = {x}^{3} + 3 {x}^{2} y + 3y {x}^{2} + {y}^{3} }}$

Take 3xy common,

$: \implies{\sf{ (x + y) ^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

Hence proved!!