Math, asked by stacygrande18, 1 month ago

If x+y+z=0, then find (x+y)³+(y+z)³+(z+x)³

Answers

Answered by MisterMega
7

Here, we have been given that x + y + z = 0, and so, to find the value of (x + y)³ + (y + z)³ + (z + x)³, follow the steps below,

 \rm  \: x + y + z = 0

And so,

 \rm x + y =  - z

 \rm \:  y + z  =  - x

 \rm \: z + x =  - y

Now, putting these values in the equation,

 \rm(x + y)³ + (y + z)³ + (z + x)³

 \to \rm  \: ( - z) ^{3}  + ( - x) ^{3}  + ( - y) ^{3}

 \to \rm \:  - z ^{3}  - x ^{3}  - y ^{3}

And hence, through the equation x + y + z = 0,

 \rm x + y + z = 0

  \to \rm \: x ^{3}  + y ^{3}  + z ^{3}  = 3xyz

And so,

 \to \rm \:   { - z}^{3}  -  {x}^{3}  -  {y}^{3}  =  - 3xyz

And hence, finally it proved that,

 \bf(x + y)³ + (y + z)³ + (z + x)³ =  - 3xyz

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Answered by mathdude500
7

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

Given that,

\sf \: x + y + z = 0  -  - (1)\\  \\

Now, we have to find the value of

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

Let assume that

\sf \: a = x + y \\  \\

\sf \: b = y + z \\  \\

\sf \: c = z + x \\  \\

So,

\sf \: a + b + c = x + y + y + z + z + x \\  \\

\sf \: a + b + c = 2(x + y +z) \\  \\

\sf \: a + b + c = 2 \times 0\\  \\

\sf \: a + b + c = 0\\  \\

\sf\implies  {a}^{3} +  {b}^{3}  +  {c}^{3}  = 3abc \\  \\

On substituting the values of a, b and c, we get

\sf\implies  {(x + y)}^{3} +  {(y + z)}^{3}  +  {(z + x)}^{3}  = 3(x + y)(y + z)(z + x) \\  \\

\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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