If $x^{2} + y^{2} + z^{2} = x y + y z + z x$ and the product of $x, y$ and $z$ is $27$ , find the value of $x^{3} + y^{3} + z^{3}$ .

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Solution

$x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$
$= (x + y + z) (x^{2} + y^{2} + z^{2} - (x^{2} + y^{2} + z^{2}))$
$= 0$

$x^{3} + y^{3} + z^{3} = 3 x y z$
$= 3 \times 27$
$= 81$

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Similar questions

If $(x^{2} + y^{2} + z^{2}) = (x y + y z + z x)$ , $what is the value of x^{3} + y^{3} + z^{3}$ ?

x + y + z = 1

x y + y z + z x = - 1

x y z = - 1

x^{3} + y^{3} + z^{3}

x, y, z \in + R

x^{2} + y^{2} + z^{2} = 27

x^{3} + y^{3} + z^{3}

x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)

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