If $x + y + z = 1, x y + y z + z x = - 1$ & $x y z = - 1$ , find $x^{3} + y^{3} + z^{3}$ .

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Solution

Given,
$x + y + z = 1$
$x y + y z + z x = - 1$
And, $x y z = - 1$
Now,
$x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$
$x^{3} + y^{3} + z^{3} + 3 = (x^{2} + y^{2} + z^{2} + 1)$
$x^{3} + y^{3} + z^{3} = x^{2} + y^{2} + z^{2} - 2$ ..........(1)
Now,
$(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x)$
$1^{2} = x^{2} + y^{2} + z^{2} - 2$
$x^{2} + y^{2} + z^{2} = 3$
Putting this value in equation 1, we get,
$x^{3} + y^{3} + z^{3} = 1$

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