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If $x+y+z = 0$ show that $x^{3}+y^{3}+z^{3}=3xyz$
Given: If $x+y+z = 0$
To find: Show that $x^{3}+y^{3}+z^{3}=3xyz$
Solution:
We know that,=$x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$
Putting $x+y+z =0$
$x^{3}+y^{3}+z^{3}-3xyz= (0)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$
$x^{3}+y^{3}+z^{3}-3xyz=0$
$x^{3}+y^{3}+z^{3}=3xyz$
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