Question

If x+y+z=0, then find (x+y)³+(y+z)³+(z+x)³
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Answer 1

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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Answer 2

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

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

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