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Question

Let $x=-12, y=7$ and $z=5$

Then $x+y+z=-12+7+5=0$

We know that if     $x+y+z=0,$ then $x^{3}+y^{3}+z^{3}=3 x y z$

$	herefore

Vedclass · Accepted Answer

$-1260$

Vedclass · Answer

Let $x=-12, y=7$ and $z=5$

Then $x+y+z=-12+7+5=0$

We know that if     $x+y+z=0,$ then $x^{3}+y^{3}+z^{3}=3 x y z$

$	herefore $      $ (-12)^{3}+(7)^{3}+(5)^{3}  =3[(-12)(7)(5)] $               $[\because(-12)+7+5=0]$

                            $ =3[-420]=-1260 $

Thus,        $ (-12)^{3}+(7)^{3}+(5)^{3}  =-1260$

Without actually calculating the cubes, find the value of each of the following : $(-12)^{3}+(7)^{3}+(5)^{3}$

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