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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?

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    $\begingroup$ You also need to stretch your interpretation to zero or negative volume when the triangle inequality is not satisfied. Gerhard "Maybe Use Hyperbolic Space Instead?" Paseman, 2018.07.19. $\endgroup$ – Gerhard Paseman Jul 19 '18 at 17:16
  • $\begingroup$ @GerhardPaseman Hyperbolic space would be interesting! But to clarify, I don't insist on a depiction valid for all possible values of $x,y,z$. The identity holds in any commutative ring... I'd be delighted with a depiction for a subset of positive real numbers. In fact, I'd even be happy with a picture for any single choice of $(x,y,z)$ (other than $x=y=z$). $\endgroup$ – Zach Teitler Jul 19 '18 at 18:15
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    $\begingroup$ The identity can be rewritten as $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(a+c)(b+c)$. Does this help? $\endgroup$ – Jairo Bochi Jul 19 '18 at 19:41
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    $\begingroup$ Can the equation $6xyz=(x+y+z)^3-(x+y)^3-(x+z)^3-(y+z)^3+x^3+y^3+z^3$ be graphically depicted? It has an obvious derivation (consider $[t^3](e^{xt}-1)(e^{yt}-1)(e^{zt}-1)$), and your identity follows from this equation by changing the signs of every pair of of variables and adding the sign-changed equations to the original equation. Does that count as a combinatorial depiction? $\endgroup$ – esg Jul 20 '18 at 19:04
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    $\begingroup$ @esg, you can visualize $(a+b)^3+(a+c)^3+(b+c)^3+6abc=(a+b+c)^3+a^3+b^3+c^3$ with the cube in my answer. $(a+b)^3$ is the left front lower 2x2 block; $(b+c)^3$ is the right back upper 2x2 block, $(a+c)^3$ is the eight corners, and $6abc$ are the six cubes on the midpoints of edges not near $a^3$ or $c^3$. Those together cover the entire cube once, and cover $a^3$, $b^3$ and $c^3$ a second time. $\endgroup$ – Matt F. Jul 22 '18 at 0:12
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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. :)

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  • $\begingroup$ I don't know what you mean by the length of a circle, and I don't know how you chop a circle into intervals. Also, I always thought a torus was $T^2$, not $T^3$. $\endgroup$ – Gerry Myerson Jul 19 '18 at 23:09
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    $\begingroup$ @GerryMyerson A circle of length $L$ is an interval of length $L$ with the extremes identified. A d-dimensional torus is a product of d circles. The torus Tˆ2 (resp. T^3) can be obtained from a square (resp. cube) by gluing the opposite sides in the most obvious way. Using this trick, it is usual to draw objects on T^2: see e.g. link Anyone able to draw (euclidian) 3-dimensional solids shouldn't have trouble with T^3. $\endgroup$ – Jairo Bochi Jul 19 '18 at 23:24
  • $\begingroup$ This is nice! I'm going to think about it—see if there's a way to describe the same thing with $(x,y,z)$, and maybe also get rid of the overlaps. Thank you! $\endgroup$ – Zach Teitler Jul 19 '18 at 23:57
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blocks

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.

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  • $\begingroup$ Thanks! This is a nice picture. If I could accept both... $\endgroup$ – Zach Teitler Jul 22 '18 at 0:36

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