TITLE: the cube of integer can be written as the difference of two square
QUESTION [10 upvotes]: This Exercise $4$, page 7, from Burton's book Elementary Number Theory.
Prove that the cube of any integer can be written as the difference of two squares. [Hint: Notice that $n^{3}=(1^{3}+2^{3}+\cdots+n^{3})-(1^{3}+2^{3}+\cdots+(n-1)^{3}).$]
Is there a way to prove that in a more natural way?
I would appreciate your help.

REPLY [3 votes]: Since the sum of the cube of the first n integers is the square of the sum of the first n integers, or:  
$\sum_{i=1}^n i^3 = \left(\sum_{i=1}^n i \right) ^2$
then:
$n^3 = \sum_{i=1}^n i^3 - \sum_{i=1}^{n-1} i^3 = \left(\sum_{i=1}^n i \right)^2 - \left(\sum_{i=1}^{n-1} i\right)^2$