Proving any integer cube is equal to a difference of two squares I know this problem has been asked before by someone. However, my problem is a bit different. I found the following problem in Elementary number theory by Burton:

Prove that the cube of any integer can be written as the difference of two squares. Notice that $n^3 = (1^3+2^3+3^3+.......+n^3) - (1^3+2^3+3^3+.......+(n-1)^3).$

My question: can this be done using the induction method? I know that:
$n^3 = (\frac{n(n+1)}{2})^2 - (\frac{(n-1)(n)}{2})^2$
I am stuck in the induction step. Can someone help?

Well, what happens when you try?
$(n+1)^3 = n^3 + 3n^2 + 3n + 1 =$
$(\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2 + 3n^2 + 3n + 1$
Meanwhile $(\frac{(n+1)(n+2)}2)^2 - (\frac {(n+1)n}2)^2 =$
$(\frac{(n+1)n}2 + \frac {2(n+1)}{2})^2 - (\frac {n(n-1)}2 + \frac {2n}{2})^2=$
$(\frac {(n+1)n}2)^2 + (n+1)^2n +(n+1)^2 - (\frac {n(n-1)}2)^2 -n^2(n-1) - n^2=$
$(\frac {(n+1)n}2)^2-  (\frac {n(n-1)}2)^2 + (n+1)^2(n+1) - n^2(n)$
$(\frac {(n+1)n}2)^2-  (\frac {n(n-1)}2)^2 + (n+1)^3 - n^3$.
$(\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2 + n^3 +3n^2 + 3n + 1- n^3=$
$(\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2 + +3n^2 + 3n + 1$.
So that's that.
