The product $(x + y + z) [{(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2}]$ is equal to

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

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

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3 x y z

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

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Solution

The correct option is D $2 [(x^{3} + y^{3} + z^{3}) - 3 x y z]$
$(x + y + z) [{(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2}]$
= $(x + y + z) [x^{2} - 2 x y + y^{2} + y^{2} - 2 y z + z^{2} + z^{2} - 2 x z + x^{2}]$
= $(x + y + z) (2 x^{2} + 2 y^{2} + 2 z^{2} - 2 x y - 2 y z - 2 z x)$
= $2 (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$
= $2 [(x^{3} + y^{3} + z^{3}) - 3 x y z]$

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Similar questions

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}]

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

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