The Binomial Cube in Four Families

Four kinds of contribution

Start with four units along every edge of the central cube, and add two in each direction on one side. The completed side is six, so the volume is two hundred sixteen. Open the pieces and notice that their shapes fall into four families.

The center is four by four by four. Its sixty-four units belong to the original cube. All three dimensions come from the old length, which is why this term uses that length three times.

A face slab has two old dimensions and one new dimension. Here each slab is four by four by two, or thirty-two. Three slabs contribute ninety-six. Their breadth comes from the old cube and their thickness from the addition.

An edge bar has one old dimension and two new dimensions. Each is four by two by two, or sixteen. Three bars contribute forty-eight. Its cross-section remembers the added thickness twice.

The last corner uses the new thickness in every direction. Two by two by two gives eight. The four families contribute sixty-four, ninety-six, forty-eight, and eight, which total two hundred sixteen.

Read dimensions instead of memorizing terms

Join the six-cube. The separate arithmetic describes one complete solid, with no missing bars or corners. Now we will use a smaller example to name the same dimensional roles.

Now see five cubed as three plus two in each direction. The original three-cube contains twenty-seven units. Two more along each edge completes a five-cube. We use this smaller example to inspect thickness, not as a substitute for the larger structures already seen.

Each face piece is three by three by two. Nine times two is eighteen. Three face pieces contribute fifty-four. The thickness multiplies the original square face; it does not replace either of the face dimensions.

Each edge piece is three by two by two. Its two-by-two cross-section contains four units, and three such cross-sections run along its length. Twelve in each bar, three bars, thirty-six altogether.

The corner is two by two by two, containing eight. Now read twenty-seven plus fifty-four plus thirty-six plus eight. The sum is one hundred twenty-five. Five cubed is accounted for by every visible region, including the thick edges and the thick corner.

The one-sided expansion

Keep the three-plus-two construction, but name its lengths. Let a denote the original side, here three, and b the added thickness, here two. The whole side is a plus b. The names allow the same picture to describe more than this one example.

The center contributes a cubed. Each face piece has two a dimensions and one b dimension, so the three faces contribute three a squared b. Point your attention to those dimensions in the slabs, not only to the letters.

Each edge piece has one a dimension and two b dimensions. Three edges contribute three a b squared. The corner has three b dimensions and contributes b cubed. That gives all four terms of the cubic expansion.

The coefficients one, three, three, one count kinds of pieces in one-sided growth. The powers describe their dimensions. If the formula becomes hard to remember, reconstruct the center, faces, edges, and corner. The representation contains the reason.

The centered expansion

Keep the same five-unit exterior, but move the cuts to put one unit on each side of the central three. The quantity remains one hundred twenty-five. Our description of the addition changes from two on one side to one on both sides.

Open this centered construction. Six face pieces each have area nine and thickness one, giving fifty-four. Twelve bars of length three give thirty-six. Eight unit corners give eight. Together with twenty-seven in the middle, the volume is still one hundred twenty-five.

For a central side a and thickness b on both ends, the whole side is a plus two b. Its volume is a cubed, plus six a squared b, plus twelve a b squared, plus eight b cubed. These coefficients count all faces, edges, and vertices.

Finish by comparing the two ways of describing growth. One-sided and centered partitions tell us which regions to count. Successive-direction growth tells us when each region enters. In every case, the dimensions of the actual pieces determine the arithmetic.

Swap the old length and the added length

Use a central side of three and an added length of two on one side. The completed side is five. Open its families: twenty-seven in the center, fifty-four in the face pieces, thirty-six in the bars and eight in the corner. Their total is one hundred twenty-five.

Join the five-cube before changing its partition. The names old length and added length have assigned different roles to three and two. The final outside length depends on their sum, but the four internal contributions remember which length was assigned to which role.

Now use a central side of two with an addition of three. The outside remains five. Opening this partition reveals eight in the center, thirty-six in the face family, fifty-four in the bars and twenty-seven in the corner. The same total has a different distribution among the four roles.

Look at the new corner, a three-cube of volume twenty-seven. That was the center contribution in the previous partition. Swapping the lengths reverses the sequence of contributions: center and corner exchange values, and the face and edge totals exchange values. The symmetry of the expansion comes from these dimensional roles.

Double a centered shell thickness

Return to a central four-cube with one unit on both ends of every axis. The outside side is six. Open the centered partition: six face slabs contribute ninety-six, twelve bars forty-eight, and eight corners eight. The center remains sixty-four. We will keep that center fixed while doubling the thickness.

Set the thickness to two on both ends. The outside side is now eight: two plus four plus two. Each of six face pieces is four by four by two, giving thirty-two each and one hundred ninety-two together. The face contribution doubled because only its thickness changed.

The twelve bars are each four by two by two. Sixteen in each bar gives one hundred ninety-two. Their total is four times the earlier forty-eight because both short dimensions doubled. The number of bars stayed twelve; their larger volume comes from changed dimensions, not from adding extra bars.

The eight corners now each contain eight, giving sixty-four altogether. Their total multiplied by eight because all three dimensions doubled. Add the unchanged center sixty-four, faces one hundred ninety-two, bars one hundred ninety-two and corners sixty-four. The result is five hundred twelve, which is eight cubed.

Equal totals need not mean equal piece counts

Keep the centered eight-cube separated. The face family and the edge family each total one hundred ninety-two, yet there are six slabs and twelve bars. Equal family totals do not require equal numbers of pieces. Each slab contains twice the volume of one bar in this particular construction.

Approach the bars to compare their dimensions. One four-by-two-by-two bar has volume sixteen; one four-by-four-by-two slab has volume thirty-two. The equality of the family totals comes from pairing twice as many bars with half as much volume in each. It is a numerical coincidence of these chosen dimensions, not a universal identity between the families.

Change back to thickness one. Now six slabs total ninety-six, while twelve bars total forty-eight. The same counts no longer give equal family totals. The geometry tells us both how many pieces there are and how much each contains; we need both pieces of information to calculate a contribution.

Assemble the six-cube. A coefficient counts repeated locations, while a product of lengths gives the volume at each location. Keeping those two jobs separate explains why centered growth has six, twelve and eight pieces without implying that those three numbers alone are the volumes being added.

Equal outside, different internal counts

Return to six cubed as four plus two on one side. The counts are one center, three face slabs, three bars, and one corner. Six cubed still equals two hundred sixteen.

Join that six-cube before changing its cuts. Keep its total volume and all three outer dimensions fixed. The next operation changes the partition, not the quantity.

Move the cuts so one unit lies on both sides of the central four. Every edge is one plus four plus one, still six. Watch the partition move inside the same outside cube.

Open the centered partition. Six face slabs each contain sixteen, twelve bars each contain four, and eight corners each contain one. The totals are again ninety-six, forty-eight, and eight. The pieces changed count and size together.

Bring the pieces together. One-sided growth counts three, three, one outside the center. Centered growth counts six, twelve, eight. Both follow from where the additional length goes; neither list is a formula detached from its cube.