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

Solution:

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

Let $x=-12, y=7$, and $z=5$

It can be observed that,

$x+y+z=-12+7+5=0$

It is known that if $x+y+z=0$, then

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

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

$=-1260$

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

Let $x=28, y=-15$, and $z=-13$

It can be observed that,

$x+y+z=28+(-15)+(-13)=28-28=0$

It is known that if $x+y+z=0$, then

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

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

$=16380$