Question

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

Answer 1

Let us consider the problem

(i) using $a^3+b^3+c^3=3abc$

since, $\left(a+b+c\right)=0$

Given: $a=28,b=-15$ and $c=-13$

hence,

${\left(28\right)}^3+{\left(-15\right)}^3+{\left(-13\right)}^2=3\times 28\times \left(-15\right)\times \left(-13\right)$

$=16380$

(ii) using $a^3+b^3+c^3=3abc$

since, $\left(a+b+c\right)=0$

Given: $a=-12,b=7$ and $c=5$

hence,

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

$=-1260$