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#
(#color(red)(Note:#Assuming -15 and 17 are included)

As the cube of a negative number is negative,
#(-15)^3# and #15^3# would cancel each other out;
#(-14)^3# and #14^3# 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#)