$1^3 + \dotsb + n^3 = (1 + \dotsb + n)^2$: reason?

Question

We have $$ 1^3 + \dotsb + n^3 = (1 + \dotsb + n)^2 $$ as we can establish by induction. But why does this hold? Can we connect it to something else?

Answer 1

Meanwhile, it generalizes to Liouville's $$ \sum_{k | n} \left( d(k) \right)^3 = \left( \sum_{k | n} d(k) \right)^2 $$

Here $d(k)$ is the number of divisors of a positive integer, with $d(1)=1.$ For a prime $p,$ we get $$ d(p^w) = w+1. $$

The identity works because it works for a prime power, that is what the original summation formula shows. Next, both sides are number theoretic "multiplicative." A multiplicative function $f(n)$ is one that applies to integers, and which has this condition: whenever $\gcd(a,b) = 1,$ we have $f(ab) = f(a) f(b).$ Any multiplicative function is completely determined by its values on prime powers. Oh, if $f(n)$ is a multiplicative function, then $$ g(n) = \sum_{k|n} f(k) $$ is also multiplicative. That requires a little proof, double sum sort of thing.

Answer 2

There's a famous proof by C. Wheatstone. http://en.wikipedia.org/wiki/Squared_triangular_number

Answer 3

Funnily, we also have

$$\int_0^x t^3\: dt = \left(\int_0^x t\: dt\right)^2.$$