Can you tell me about the (probably) well known relationship between the coefficients of a cubic and some features of a rectangular solid?

Question

If we look at the expansion of this $$(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ca)x+abc$$ And consider a rectangular solid with length, width and height of $a, b, c$ respectively.

Then $$l_{edges}=4(a+b+c)$$ $$A_{surface}=2(ab+bc+ca)$$ $$V=abc$$

Which match up very nicely with the terms of this cubic. (Well, three of the terms, anyway.) Especially if we set $x=2$.

After I thought about this (daydreaming in a meeting at the end of the day) I wondered if this was something everyone knew, but I had just missed out on. And then I wondered if there was anything fun that could be done with this.

I would love some leads on key words (so I can find an article addressing this and see what paths this leads to) or some things to explore that are connected to this.

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Okay, the obvious paths are the 2nd power case power case. $$(x+a)(x+b)=x^2+(a+b)x+ab$$ $$Perimeter=2(a+b)$$ $$Area=ab$$

And the 4th and higher cases. $$(x+a)(x+b)(x+c)(x+d)=x^4+(a+b+c+d)x^3+(ab+bc+cd+da)x^2+(abc+bcd+cda+dab)x+abcd$$ $$l_{edge}=8(a+b+c+d)$$ $$A_{sides}=4(ab+bc+cd+da)$$ $$V_{sides}=2(abc+bcd+cda+dab)$$ $$HyperVolume=abcd$$

Answer 1

You can think of a closed interval $[0, a]$ of length $a$ as consisting of two points (namely its endpoints $0$ and $a$) together with the open interval $(0, a)$ of length $a$. Now the set

$$[0, a] \times [0, b] \times [0, c] \subset \mathbb{R}^3$$

is a rectangular solid with side lengths $a, b, c$. Divide up each side as its two endpoints together with the open interval, then "expand out the product" and you get your result. This also explains Sammy's observation about vertices in the comments.

Answer 2

The other answer by Qiaochu is great, but I'll offer a slightly different perspective. You have scratched the surface of a rather deep topic. We can think of the polynomial as a (finite) generating function that keeps track of the measurements of the rectangular faces of each dimension.

As you observed, there's something special about $x=2$, so let's fix that at the start. (We can relax this later.) Let's look at an $n$-dimensional rectangular prism. We will denote the lengths using subscripts—so, e.g., $a_1, a_2, a_3$ rather than $a, b, c$ in dimension $n=3$—and consider the polynomial $$ p_n(a_1, a_2, \dots, a_n) = (2 + a_1) (2 + a_2) \cdots (2 + a_n) $$ This polynomial of total degree $n$ in $n$ variables is our generating function.

If we expand it, then the degree $d$ term measures the total "volume" of the $d$-dimensional faces of the prism, where as usual "volume" means
\begin{array}{rcl} d=0 &\leftrightarrow &\text{vertex count ($0$-faces)} \\ d=1 &\leftrightarrow &\text{length of edges ($1$-faces)} \\ d=2 &\leftrightarrow &\text{area of $2$-faces} \\ d=3 &\leftrightarrow &\text{volume of $3$-faces} \\ &\vdots & \\ d=n &\leftrightarrow &\text{hypervolume of entire $n$-prism} \end{array}

For example, with $n=3$, your observation in the post looks like \begin{align} p_3(a_1, a_2, a_3) &= (2 + a_1) (2 + a_2) (2 + a_3) \\ &= 8 + 4(a_1 + a_2 + a_3) + 2(a_1a_2 + a_1a_3 + a_2a_3) + 1(a_1a_2a_3), \end{align} in this notation. And if we want to measure, say, the total surface area, we look at the degree $d = 2$ term $$ 2(a_1a_2 + a_1a_3 + a_2a_3) $$ and treat the variables as actual length measurements.

But, why does this work? We can build up the formula by induction. Given an $n$-prism of dimension $a_1 \times a_2 \times \dots \times a_n$, we construct a prism of dimension one bigger by making $2$ copies of the $n$-prism, and separating them by a new distance $a_{n+1}$. In the process each existing $d$-face is now extruded in the new dimension to become a new $(d+1)$-face. This is kept track of perfectly by \begin{align} p_{n+1}(a_1, a_2, \dots, a_n, a_{n+1}) &= 2 \cdot p_n(a_1, a_2, \dots, a_n) + a_{n+1} \cdot p_n(a_1, a_2, \dots, a_n) \\ &= (2 + a_{n+1}) \cdot p_n(a_1, a_2, \dots, a_n) \\ &= (2 + a_1) (2 + a_2) \cdots (2 + a_n) (2 + a_{n+1}) \end{align}

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If we just want to count how many $d$-faces are in the $n$-prism, we specialize (construct a polynomial of just one variable): $$ q_n(a) = p_n(a, a, \dots, a) = (2 + a) (2 + a) \cdots (2 + a) = (2 + a)^n, $$ which by the Binomial Theorem looks like $$ \sum_{d=0}^n \binom{n}{d} 2^{n-d}a^d, $$ showing us, by reading off the coefficient in degree $d$, that there are precisely $$ 2^{n-d} \binom{n}{d} $$ faces of dimension $d$ in a rectangular $n$-prism.

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Here are two interesting generalizations to explore:

1. Find a family of $n$-dimensional objects for which $(3 + a_1) (3 + a_2) \cdots (3 + a_n)$ generates its measurements.

Hint: think of a $2 \times 2$ four square court for dimension $n=2$. Can you generalize this?

2. Find a family of $n$-dimensional objects for which $(1 + a)^{n+1}$ counts its faces in each dimension.

Hint: a tetrahedron has $4$ vertices, $6$ edges, $4$ faces, and $1$ volume.