Question

Without actually calculating the cubes, find the value of ${(-12)}^3+{\left(7\right)}^3+{\left(5\right)}^3$

• A. 0
• B. -36
• C. -1620
• D. -1260

Answer 1

The correct option is D

-1260

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

When $(x+y+z)=0$

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

$-12+7+5=0$

$\therefore {(-12)}^3+{\left(7\right)}^3+{\left(5\right)}^3=3(-12)\left(7\right)\left(5\right)=-1260$