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

Apr 21, 2015

The sum of the cubes would look like this:

${\left(- 15\right)}^{3} + {\left(- 14\right)}^{3} + . . + {\left(- 1\right)}^{3} + {0}^{3} + {1}^{2} + . {.14}^{3} + {15}^{3} + {16}^{3} + {17}^{3}$
(color(red)(Note:Assuming -15 and 17 are included)

As the cube of a negative number is negative,
${\left(- 15\right)}^{3}$ and ${15}^{3}$ would cancel each other out;
${\left(- 14\right)}^{3}$ and ${14}^{3}$ would cancel each other out;
and so on...

$\cancel{{\left(- 15\right)}^{3}} + \cancel{{\left(- 14\right)}^{3}} + . . + \cancel{{\left(- 1\right)}^{3}} + {0}^{3} + \cancel{{1}^{2}} + . . \cancel{{14}^{3}} + \cancel{{15}^{3}} + {16}^{3} + {17}^{3}$

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)

$= \left(16 + 17\right) \left({16}^{2} - \left(16 \cdot 17\right) + {17}^{2}\right)$

$= 33 \cdot \left(256 - 272 + 289\right)$

$= 33 \cdot 273$

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)