I'm looking for a geometric or combinatorial depiction of the algebraic identity
$$
 xyz = \frac{1}{24} \Big\{(x+y+z)^3 - (x-y+z)^3 - (x+y-z)^3 + (x-y-z)^3 \Big\}.
 \label{\*}\tag{$\*$}
$$
Here is the kind of thing I'd like. For the simpler identity $xy = \frac{1}{4} \big\{(x+y)^2 - (x-y)^2 \big\}$ we can rearrange to $(x+y)^2 = (x-y)^2 + 4xy$. Now, if $x>y>0$, we can take a square with side length $x-y$, and $4$ rectangles of size $x \times y$, and put them together to make a square of side length $x+y$. Just put the little square in the middle and the rectangles around its sides.


My idea was to rearrange $\eqref{\*}$ into
$$
 (x+y+z)^3 = (-x+y+z)^3 + (x-y+z)^3 + (x+y-z)^3 + 24xyz .
$$
Then, suppose $x,y,z>0$ and they satisfy triangle inequalities. Now three cubes of edge lengths $-x+y+z$, $x-y+z$, and $x+y-z$, plus $24$ "bricks" of size $x \times y \times z$, have the same volume as a cube of edge length $x+y+z$. Unfortunately it's not generally possible to stack the 3 little cubes plus $24$ bricks into a big cube.


(Try $(x,y,z)=(11,13,17)$. The only way to get the right areas of faces of the big cube is for each face of the big cube to have exactly one face of a little cube, plus $4$ faces of bricks. And the little cubes have to be centered on the big cube faces; they can't be in the corners or the middles of the edges. But there are $6$ big cube faces and only $3$ little cubes.)


This is a bit open-ended, but can anyone suggest a different way to illustrate the identity, especially if it can be depicted in a graphic? Maybe a different algebraic rearrangement of $\eqref{\*}$, or another shape besides cubes?



Geometric/combinatorial depiction of algebraic identity?
The identity can be rewritten as



> 
> $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(a+c)(b+c)$
> 
> 
> 


by means of a linear change of variables $a:=(−x+y+z)/2$, etc.


Let $T$ be a circle of length $a+b+c$, and let's chop it into three intervals $A$, $B$, $C$ of respective lengths $a$, $b$, $c$. Consider also the intervals $A':=B \cup C$, $B':=A \cup C$, $C':=A \cup B$.


In the torus $T^3$, let's consider two types of bricks: **small** ones with sides of lengths $a$, $b$, $c$, and **large** ones, with sides of lengths $a+b$, $a+c$, $b+c$.


Consider the following large bricks:



> 
> $L\_1 := A' \times B' \times C'$,
> 
> 
> $L\_2 := B' \times C' \times A'$,
> 
> 
> $L\_3 := C' \times A' \times B'$,
> 
> 
> 


and let $U$ be their union.
Note (\*): the triple intersection is empty, and each pairwise intersection is a small brick (e.g., $L\_1 \cap L\_2 = C \times A \times B$).


Next, we can re-assemble the torus $T^3$ using the following (essentially disjoint) pieces:


* the three cubes $A \times A \times A$, $B \times B \times B$, and $C \times C \times C$;
* the solid $U$; and
* the three small bricks $A \times C \times B$, $C \times B \times A$, and $B \times C \times A$.


By the previous note (\*), the union of $U$ with these three small bricks has the volume of $3$ large bricks. So we obtain the desired inequality.


To be honest, I have trouble in visualizing all of this simultaneously, but it should be possible. :)


