# 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. Jul 19, 2018 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 $x=y=z$). Jul 19, 2018 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? Jul 19, 2018 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, 2018 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. Jul 22, 2018 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 $T^3$. Jul 19, 2018 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. Jul 19, 2018 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! Jul 19, 2018 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... Jul 22, 2018 at 0:36