# 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?

• 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. – Gerhard Paseman Jul 19 '18 at 17:16
• @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 $x=y=z$). – Zach Teitler Jul 19 '18 at 18:15
• 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? – Jairo Bochi Jul 19 '18 at 19:41
• 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? – esg Jul 20 '18 at 19:04
• @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. – Matt F. Jul 22 '18 at 0:12

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. :)

• 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 $T^3$. – Gerry Myerson Jul 19 '18 at 23:09
• @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. – Jairo Bochi Jul 19 '18 at 23:24
• 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! – Zach Teitler Jul 19 '18 at 23:57 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.

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