Question

Question 14
Without actually calculating the cubes, find the value of each of the following:
$\left(i\right)(-12{)}^3+(7{)}^3+(5{)}^3$
$\left(ii\right)(28{)}^3+(-15{)}^3+(-13{)}^3$

Answer 1

From the above question, we have $x^3+y^3+z^3=3xyz,ifx+y+z=0$.
(i) Here - 12 + 7 + 5 = 0
So,
$\Rightarrow (-12{)}^3+(7{)}^3+(5{)}^3=3(-12\left)\right(7\left)\right(5)=-1260$

(ii) Here 28 + (-15) + (-13) = 0
So,
$\Rightarrow (28{)}^3+(-15{)}^3+(-13{)}^3=3\times 28(-15\left)\right(-13)=16380$