How do you find the sum of the cubes of the integers in the interval -15 and 17?

1 Answer
Apr 21, 2015

The sum of the cubes would look like this:

(-15)^3 + (-14)^3+..+(-1)^3 + 0^3 + 1^2 + ..14^3 + 15^3 + 16^3 + 17^3(15)3+(14)3+..+(1)3+03+12+..143+153+163+173
(color(red)(Note:Note:Assuming -15 and 17 are included)

As the cube of a negative number is negative,
(-15)^3(15)3 and 15^3153 would cancel each other out;
(-14)^3(14)3 and 14^3143 would cancel each other out;
and so on...

cancel((-15)^3) + cancel((-14)^3)+..+cancel((-1)^3) + 0^3 + cancel(1^2) + ..cancel(14^3) + cancel(15^3) + 16^3 + 17^3

What would remain is :

0^3 + 16^3 + 17^3

= 16^3 + 17^3

WE know that color(blue)(a^3+b^3 = (a + b)(a^2 - ab + b^2)

= (16 + 17)(16^2 - (16*17) + 17^2)

= 33*(256 - 272 + 289)

= 33*273

color(green)( = 9009

(color(red)(Note:If -15 and 17 are not included, then the answer would be 15^3 + 16^3 = color(green)(7471)