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

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.
