Math, asked by ShivaniPurple, 1 year ago

prove that (x+y)³-(x-y)³-6y(x²-y²)=8y³

Answers

Answered by Anonymous
9
x^3 + y^3 - 3 x. y^2 -3 x^2.y - x^3 + y^3 + 3x.y^2 - 3 X^2 .Y - 6 YX^2 + 6 y^3
=8 y ^3
Answered by mathdude500
2

\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}

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