Question: Is there an intuitive way to get cubic sum? From this post: [combination of quadratic and cubic series](https://math.stackexchange.com/questions/876873/combination-of-quadratic-and-cubic-series/876875#876875) and Wikipedia: [Faulhaber formula](http://en.wikipedia.org/wiki/Faulhaber%27s_formula), I get $$1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$$
I think the cubic sum is squaring the arithmetic sum $$1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$$
But how to prove it? Please help me. Grazie!

Answer: 
Maybe this will help you visualize it:

![](https://i.stack.imgur.com/MXOt5.png)

[Source.](http://en.wikipedia.org/wiki/File:Sum_of_cubes.png)

or this one which is clearer:

$\phantom{XXXXXXXX}$![e](https://i.stack.imgur.com/MjQGP.jpg)
Here's a proof by induction
I came up with
about two years ago.

Let
$s\_k(n)
=\sum\_{i=1}^n i^k
$.
Want to show that
$s\_3(n) = (s\_1(n))^2
$.

If you want to
prove by induction
that $a(n) = b(n)$,
there are two possibilities:
(1) Show
$a(0)-b(0) = 0$
and
$a(n)-b(n) = 0
\implies
a(n+1)-b(n+1) = 0$;
(2) show
$a(0) = b(0)$
and
$a(n) = b(n)
\implies
a(n+1)-a(n)
=b(n+1)-b(n)
$.

These are,
of course,
equivalent,
but I find that (2)
is often easier
since the math
is simpler
in the initial parts.

In this case,
$s\_3(n+1)-s\_3(n)
=(n+1)^3
=n^3+3n^2+3n+1
$
and

$(s\_1(n+1))^2-(s\_1(n))^2
=(s\_1(n)+n+1)^2-(s\_1(n))^2\\
=s\_1^2(n)+2(n+1)s\_1(n)+(n+1))^2-(s\_1(n))^2\\
=2(n+1)s\_1(n)+n^2+2n+1\\
$

so we want to show that
$n^3+3n^2+3n+1
=2(n+1)s\_1(n)+n^2+2n+1
$
or
$n^3+2n^2+n
=2(n+1)s\_1(n)
$
or
$n(n^2+2n+1)
=2(n+1)s\_1(n)
$
or
$n(n+1)^2
=2(n+1)s\_1(n)
$
or
$s\_1(n)
=\dfrac{n(n+1)}{2}
$.

This can either
be assumed known
or, in turn,
proved by induction using
$\dfrac{(n+1)(n+2)}{2}
-\dfrac{n(n+1)}{2}
=\dfrac{n^2+3n+2-(n^2+n)}{2}
=\dfrac{2n+2}{2}
=n+1
$.