Geometric/combinatorial depiction of algebraic identity? 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?
 A: 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. :)
A: 
This shows the identity
$$(a+b+c)^3 = a^3 + b^3 + c^3 + 3a(a+b)(b+c) + 3c(b+c)(a+b)$$
which builds on Jairo’s answer.
Each summand represents a block in the cube, e.g. $a(a+b)(b+c)$ represents the block $(0,a) \times (0,a+b) \times (a,a+b+c)$, and multiplication by 3 represents cyclic permutation through the axes.  The cube is a sum of 9 smaller blocks, though two are not visible in the drawing.
