Math, asked by dharmenderkr1056, 7 months ago

simplify (X+Y)³-(X-Y)³-6Y(X+Y)(X-Y)​

Answers

Answered by yashika1951
7

\huge\underline\bold\red{AnsweR...}

answer in above attachment....

Step-by-step explanation:

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Answered by mathdude500
2

Answer:

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

Given expression is

\sf \: {(X + Y)}^{3} -  {(X - Y)}^{3} - 6Y( {X}^{2} -  {Y}^{2}) \\

Ler assume that X = x and Y = y

So, above expression can be rewritten

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

can be rewritten as

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

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

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

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

\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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