Question

Factorise :

( 1 ) ( x + 1 ) ³ - ( x - 1 )³
( 2 ) ( x + 1 ) ³ + ( x - 1 )³

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Answer 1

Answer:

Given : x(x+y)

3

−3x

2

y(x+y)

Taking x(x+y) as common in the given question,

we get, x(x+y)[(x+y)

2

−3xy]

We know that, (a+b)

2

=a

2

+b

2

+2ab

x(x+y)

3

−3x

2

y(x+y)=x(x+y)(x

2

+y

2

+2xy−3xy)

So we get,

x(x+y)

3

−3x

2

y(x+y)=x(x+y)(x

2

+y

2

−xy)

Step-by-step explanation:

hope it's helpful to you

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Answer 2

→ Hey Mate,

→ Given Question:-

→ Factorise :

→ ( 1 ) ( x + 1 ) ³ - ( x - 1 )³

→ ( 2 ) ( x + 1 ) ³ + ( x - 1 )³

→ Solution:-

→ (1) ( x + 1 )³ - ( x - 1 )³

→ =  { ( x + 1 ) - ( x - 1 ) } { ( x + 1 )² + ( x + 1) ( x-1 ) + (x - 1)²}

→ = ( x + 1 - x + 1 ) {( x² + 2x + 1 ) + (x² - 1) + (x²-2x + 1)}

→ = 2 ( x² + 2x + 1 + x² - 1 + x² -2x + 1)

→ = 2 ( 3x² + 1 )

→ (2) ( x + 1 ) ³ + ( x - 1 )³

→ = { ( x + 1 ) + ( x - 1) } { (x+1)²- (x+1) (x-1) + (x-1)²}

→ = (x + 1 + x-1) {( x² + 2x + 1 ) - (x²-1)+ (x²-2x+1)

→ = 2x (x² + 2x + 1 -x² + 1+x² -2x + 1)

→ = 2x  (x² + 3)

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