use induction to prove
1^{3} + 2^{3} + 3^{3} + . . . + n^{3} = (1 + 2 + 3 + . . . + n)^{2 }
I am going to assume that we are aware of the identity
$\displaystyle \sum_{i = 1}^k = \frac{k(k + 1)}{2}$
It is not necessary, but useful.
First, does this hold for n = 1? Obviously.
So assume that there exists a n = k such that
$\displaystyle 1^3 + 2^3 + ~...~ + k^3 = (1 + 2 + ~...~ + k)^2$
So what can we do with n = k + 1?
$\displaystyle (1^3 + 2^3 + ~...~ + k^3) + (k + 1)^3 = ((1 + 2 + ~...~ + k) + (k + 1))^2$
By our assumption the first k terms on the LHS can be replaced by ( )^2. We are also going to expand the RHS as a square:
$\displaystyle (1 + 2 + ~...~ + k)^2 + (k + 1)^3 = (1 + 2 + ~...~ + k)^2 + 2(1 + 2 + ~...~ + k)(k + 1) + (k + 1)^2$
The first terms on both sides cancel, leaving
$\displaystyle (k + 1)^3 = 2(1 + 2 + ~...~ + k)(k + 1) + (k + 1)^2$
I'm going to leave the rest to you, but with the reminder of the sum above:
$\displaystyle \sum_{i = 1}^k = 1 + 2 + ~...~ + k = \frac{k(k + 1)}{2}$
If you need clarification, please let us know.
-Dan