Math, asked by arnavguglani05, 11 months ago

If (x + y)^3 - (x – y)^3 – 6y (x^2 - y^2)= ky^2, then find k

Answers

Answered by mathdude500
3

Answer:

\boxed{\sf \: k = 8y \: } \\  \\

Step-by-step explanation:

Given that,

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

can be rewritten as

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

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

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

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

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

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

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} = k {y}^{2}  \\

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

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

\sf \:  {8y}^{3} = k {y}^{2}  \\

\implies\sf \: k = 8y \\  \\

\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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