Question

without actually computing the cubes evaluate 30^3-18^3-12^3

Answer 1

If
$30 - 18 - 12 = 0 \\ or \: 30 - 30 = 0$
Then using the formula
$a {}^{3} + b {}^{3} + c {}^{3} = 3abc$
Let a =30
b = -18
c= -12
So,
$(30) {}^{3} + ( - 18) {}^{3} + ( - 12) {}^{3} \\ = 3 \times 30 \times - 18 \times - 12 \\ = 19440$

Answer 2

30^3-18^3-12^3= 30^3+(-18^3)+(-12^3)
let a= 30,b=-18,c=-12
a+b+c=30+(-18)+(-12)=0
If a+b+c=0
Then a^3+b^3+c^3=3abc
=3×30×(-18)×(-12)
=19440