If $x^{2} + y^{2} + z^{2} = 2 (x + z - 1),$ then what is the value of $x^{3} + y^{3} + z^{3} = ?$

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Solution

The correct option is C
2

$x^{2} + y^{2} + z^{2} = 2 (x + z - 1) x^{2} + y^{2} + z^{2} = 2 x + 2 z - 2 x^{2} - 2 x + 1 + y^{2} + z^{2} - 2 z + 1 = 0 (x - 1)^{2} + y^{2} + (z - 1)^{2} = 0$
From the above equation it is clear that
x=1, y=0, z=1
Substituting these in
$x^{3} + y^{3} + z^{3} = 1^{3} + 0^{3} + 1^{3} = 2$

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

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x^{3} + y^{3} + z^{3} - 3 x y z = \frac{1}{2} (x + y + z) [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]

Verify that : $x^{3} + y^{3} + z^{3} - 3 x y z = \frac{1}{2} (x + y + z) [{(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2})]$

x^{3} + y^{3} + z^{3} - 3 x y z = \frac{1}{2} (x + y + z) [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]

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