# Without actually calculating the cubes, find the value of each of the following:(i) $(-12)^{3}+(7)^{3}+(5)^{3}$(ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$

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To do:

We have to find the values of each of the given expressions.

Solution:

We know that,

$x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$

If $x+y+z =0$, then

$x^{3}+y^{3}+z^{3}-3xyz= (0)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$

$x^{3}+y^{3}+z^{3}-3xyz=0$

$x^{3}+y^{3}+z^{3}=3xyz$

(i) $(-12)^{3}+(7)^{3}+(5)^{3}$

Here,

$a=-12, b=7, c=5$

This implies,

$a+b+c=-12+7+5$

$=-12+12$

$=0$

Therefore,

$(-12)^{3}+(7)^{3}+(5)^{3}=3(-12)(7)(5)$

$=-36\times35$

$=-1260$

(ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$

$a=28, b=-15, c=-13$

This implies,

$a+b+c=28+(-15)+(-13)$

$=28-(15+13)$

$=28-28$

$=0$

Therefore,

$(28)^{3}+(-15)^{3}+(-13)^{3}=3(28)(-15)(-13)$

$=84\times195$

$=16380$

Updated on 10-Oct-2022 13:39:07