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 are some *'Proof without Words'* for this identity;

As Hakim showed (but this one might be slightly clearer);
[![#1](https://i.stack.imgur.com/GMx5N.png)](https://i.stack.imgur.com/GMx5N.png)

Here's another very clean illustration by Anders Kaseorg;
[![#2](https://i.stack.imgur.com/P8ywd.png)](https://i.stack.imgur.com/P8ywd.png)

Here, the total top volume is undoubtedly $1^2+2^2+3^3+\cdots +n^2,$ while on the volume of the bottom part is;
$$1\left(1+2+\cdots+n\right)+2\left(1+2+\cdots+n\right)+\cdots +n\left(1+2+\cdots+n\right)=\left(1+2+\cdots+n\right)^2.$$