I tried the same for (a+b)[sup]3[/sup]
Here, of course, instead of a square, we have to use a cube. And sure enough, the total volume (instead of area) of the cube comes to be
(a+b)[sup]3[/sup] = a[sup]3[/sup] + b[sup]3[/sup] + 3a[sup]2[/sup]b + 3ab[sup]2[/sup]
You can also use it to get the formula of squares and cubes of differences, ie., (a-b)[sup]2[/sup] and (a-b)[sup]3[/sup]. Just that, in the square or the cube, instead of having the total length as (a+b), keep the total length as a, and have another segment b in it. Compute the required area or volume, and it comes to be correct.
Now, this will prove to be a little bit more difficult if we go to more number of variables, like (a+b+c)[sup]2[/sup] or (a+b+c)[sup]3[/sup], as we will have more rectangles or more cubes to compute the area or volume.
Also, when the power goes beyond 3, we will have to imagine a <a href="https://en.wikipedia.org/wiki/Tesseract" target="_blank" rel="nofollow noopener noreferrer">Tesseract</a> or the appropriate hypercube, and it becomes impractical after that..
