Visualizing $3^3+4^3+5^3=6^3$

Question

I was wondering how would the visualization of $3^3+4^3+5^3=6^3$ look? I tried some 3D sketches but when calculating none work. Any ideas would be highly appreciated.

Answer 1

$$ 5^3 + 3^3 + 4^3 = 5^3 + 3 \cdot (3^2 + 4^2) + 4^2 = 5^3 + 3 \cdot 5^2 + 3 \cdot 5 + 1 = (5 + 1)^3 $$ So a cube of side $6$ units can be made by taking a $\color{green}{\text{cube of side 5 units} }$, adding $ \color{blue}{\text{3 5-by-5 squares}}$ at the top, right and left, then adding $\color{red}{\text{3 lines of length 5}}$ along $3$ edges and finally adding $\color{black}{ \text{1 cube} }$ at the corner.

Source of picture of cube

Answer 2

You can think to a big cube with side $6$ made up of $6^3=216$ unitary small cubes that you can take to build up exactly three small cubes wit sides $3$,$4$, and $5$.

Answer 3

There is a 3D dissection on Wikipedia https://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Mathematics/2018_March_17#/media/File:Plato_number.svg - but it is not a very "natural" dissection.

Answer 4

No complete answer but some thoughts:

Subtracting a 5-cube from a 6-cube leaves

$$6^3-5^3=216-5^3=91.$$

Imagine you place the 5-cube into the 6-cube and check what is left over. Lets assume we rotate it so that the 5-square "on ground" is inscribed into the 6-square.

Now we extend the 5-cube by one unit to the top and subtract that as well, thats

$$91-5^2=91-(3^2+4^2)=66.$$

We take these additional 25 volume units we have subtracted from one layer of the 3-cube and one layer of the 4-cube, thats Pythagoras and that can be visualised convincingly. Now we are left with showing

$$3\cdot3\cdot2 + 4\cdot4\cdot3=66.$$

$3\cdot3\cdot2$ and $4\cdot4\cdot3$ is what is left over from the 3- and 4-cube and 66 is apprently the volume of the remaining 4 trigonal prisms (="wedges" and when they are inscribed as suggested wil have sidelengths $3+\sqrt{\frac 72}$, $3-\sqrt{\frac 72}$ and 6 each).

So what remains to show is $$3\cdot3\cdot2 + 4\cdot4\cdot3=66,$$

Since the wedges have hight $6$ and both remainders of the smaller cubes contain 6 as a factor we could divide each figure by $6$ getting four bodies of height 1, and then we can collapse the bodies to 2D polygons arriving at four area $\frac{11}{4}$ triangles each and rectangles with sides $3\times4$, and $4\times2$, representing the equation

$$3 \cdot 1 + 4\cdot 2 = 4 \frac12 \bigg[3-\sqrt{\frac 72}\bigg]\cdot\bigg[3+\sqrt{\frac 72}\bigg]=11$$

I did not come up with a nice diagram to show that but maybe it can be somehow derived from a construction of $\sqrt{\frac{7}{2}}$, somehow in the style of the linked Pythagoras proofs.