Question

Simplify (x+y)³ -(x-y)³-6y(x²-y²)

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Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\sf \: {(x + y)}^{3} - {(x - y)}^{3} - 6y( {x}^{2} - {y}^{2})$

can be rewritten as

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

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

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

$\sf \:={(x + y)}^{3} - {(x - y)}^{3} - \red{3[ (x + y) - (x - y)]}( {x}^{2} - {y}^{2})$

$\sf \: = {(x + y)}^{3} - {(x - y)}^{3} - {3( {x}^{2} - {y}^{2}) [ (x + y) - (x - y)]}$

$\sf \: = {(x + y)}^{3} - {(x - y)}^{3} - {3(x + y)(x - y)[ (x + y) - (x - y)]}$

We know,

$\boxed{\sf \: {(a - b)}^{3} = {a}^{3} - {b}^{3} - 3ab(a - b) \: }$

So, using this algebraic identity, we get

$\sf \:={[ x + y - (x - y)]}^{3}$

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

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

$\sf \: = {8y}^{3}$

$\implies\sf \: {(x + y)}^{3} - {(x - y)}^{3} - 6y( {x}^{2} - {y}^{2}) = 8y^3$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

=(x+y)

3

−(x−y)

3

−6y(x

2

−y

2

)

=(x+y)

3

−(x−y)

3

−6y(x−y)(x+y)

=(x+y)

3

−(x−y)((x−y)

2

+6y(x+y))

=(x+y)

3

−(x−y)(x

2

+y

2

−2xy+6xy+6y

2

)

=(x+y)

3

−(x−y)(x

2

+7y

2

+4xy)

=x

3

+y

3

+3x

2

y+3xy

2

−x

3

−7xy

2

−4x

2

y+x

2

y+7y

3

+4xy

2

=8y

3