In short

Every cube identity is the same idea as the square one, with one extra axis multiplying through. (a+b)^2 slices a square into 2^2 = 4 tiles; (a+b)^3 slices a cube into 2^3 = 8 boxes. The general (a+b)^n slices an n-dimensional hypercube into 2^n pieces. Pascal's triangle is just the bookkeeping: row n counts how many pieces of each type the dissection has, and the recurrence "each entry = sum of the two above" is what happens when you grow the cube by one dimension. Why 2^n? Each axis splits into two pieces (a or b), the n axes are independent, and 2 \times 2 \times \dots \times 2 = 2^n.

You have already seen the square dissection — four tiles whose areas add to (a+b)^2. You have seen the cube dissection — eight boxes whose volumes add to (a+b)^3. The natural question, the one that makes the whole picture click: how are these two pictures the same picture? Where exactly does the extra dimension come from, and why does Pascal's triangle suddenly start describing cubes?

The answer is a single move you can do with your hands: take the 2D picture and extrude it. Push every flat tile straight back along a new third axis, and the 4 tiles become 4 \times 2 = 8 boxes — exactly the eight pieces of the cube identity. Do it again into 4D, and the 8 becomes 16. The bridge from squares to cubes is one extrusion, and Pascal's triangle is what tracks the bookkeeping.

Squares as 2D, cubes as 3D

A square of side a + b has area (a+b)^2. One horizontal cut and one vertical cut — two cuts total — slice it into 2 \times 2 = 4 tiles: an a^2 corner, a b^2 corner, and two a \times b rectangles. That is the entire (a+b)^2 identity, sitting flat on the page.

A cube of side a + b has volume (a+b)^3. Three perpendicular planar cuts — one per axis — slice it into 2 \times 2 \times 2 = 8 boxes: an a^3 corner, a b^3 corner, three a^2 b slabs (one per axis), three a b^2 slabs (one per axis). That is the entire (a+b)^3 identity.

The pattern continues. A 4D hypercube of side a + b has hyper-volume (a+b)^4. Four perpendicular hyperplane cuts slice it into 2^4 = 16 rectangular pieces. You cannot draw it, but the counting still works: 1 + 4 + 6 + 4 + 1 = 16. In general, (a+b)^n is an n-dimensional hypercube cut into 2^n rectangular pieces, and you can name each piece by an n-letter string of as and bs — one letter per axis. Why 2^n pieces? Each of the n axes contributes a binary choice ("is this piece's edge on this axis a or b?"), and the n choices are independent. The total is 2 \times 2 \times \dots \times 2 = 2^n.

Bridge between the square dissection and the cube dissection by extrusionTop: a square of side a plus b, cut into four tiles labelled a squared, ab, ab, b squared. Bottom: an isometric cube of side a plus b, cut into eight boxes. Arrows from each square tile point downward, showing that each 2D tile extrudes into two 3D boxes when a third axis is added — one box gets length a along the new axis, the other gets length b. 2D: square cut into 2² = 4 tiles a² ab ab b² a b a b extrude × a extrude × b extrude 3D: each tile extrudes into 2 boxes → 4 × 2 = 8 a³ a²b a²b ab² a²b ab² ab² b³ a³ + 3a²b + 3ab² + b³
The square dissection extruded into the cube dissection. Each of the four 2D tiles becomes a pair of 3D boxes when you push it along a new third axis — once with the new edge length $a$, once with the new edge length $b$. Four tiles × two extrusions = eight boxes. The $a^2$ tile gives $a^3$ and $a^2 b$; each $ab$ tile gives $a^2 b$ and $a b^2$; the $b^2$ tile gives $a b^2$ and $b^3$.

Pascal's row n gives "how many of each type"

Stack the type counts for each n:

This is Pascal's triangle, but you have not had to memorise it — you have read it off the dissection. The k-th entry in row n is the number of n-letter strings with exactly k b's, i.e. \binom{n}{k}. Geometrically it is the number of pieces in the n-cube whose edges are b on exactly k axes.

Look at the 3 in row 3. It is also equal to 1 + 2 — the 1 and 2 sitting in row 2 just above it. Why? Because of how the extrusion works. The three a^2 b slabs in the cube come from two distinct sources: (a) the single a^2 tile in the square, extruded along the new b-direction (one slab), plus (b) the two ab tiles in the square, each extruded along the new a-direction (two slabs). One plus two equals three. The Pascal recurrence \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} is literally the geometric merge.

The double-counting is not a coincidence. Each new dimension you add gives every old piece two children — one with the new edge equal to a, one with the new edge equal to b. Pieces of the same type in the new dimension can come from two old types, and Pascal's row records the merge.

The recurrence as physical addition

Here is the bookkeeping written out. To go from row n - 1 to row n, you add a new axis. Every (n-1)-piece is "reborn" twice in the n-cube: once with the new edge a (so it stays the same type, just gains an a), and once with the new edge b (so its type bumps up by one b).

If row n-1 has \binom{n-1}{k} pieces of type "k b's", then in row n:

A type-"k b's" piece in row n is therefore the sum of: pieces of type "k" from row n-1 (the new axis is a) plus pieces of type "k-1" from row n-1 (the new axis is b). That is

\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}

and it is exactly Pascal's recurrence. The recurrence is not a rule you have to memorise — it is the statement "when you grow the cube by one dimension, two old groups merge into one new group". Why "double-count and merge"? Because the new edge can be either a or b, and adjacent groups (those differing by one b) end up overlapping in the new row.

Worked examples

Example 1 — derive $(a+b)^3$ by multiplying by $(a+b)$

You can also derive the cube identity algebraically by multiplying the square identity by one more factor of (a+b). This is the same extrusion, written in symbols.

(a+b)^3 = (a+b) \cdot (a+b)^2 = (a+b)(a^2 + 2ab + b^2)

Distribute the (a+b) across each term:

  • a \cdot (a^2 + 2ab + b^2) = a^3 + 2 a^2 b + a b^2
  • b \cdot (a^2 + 2ab + b^2) = a^2 b + 2 a b^2 + b^3

Add them:

a^3 + 2a^2 b + a b^2 + a^2 b + 2 a b^2 + b^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3

The two contributions 2 a^2 b + a^2 b = 3 a^2 b and a b^2 + 2 a b^2 = 3 a b^2 are the algebraic version of the geometric merge — exactly the "1 + 2 = 3" Pascal step. Multiplying by (a + b) is extruding by one new axis: the a branch stretches every existing tile by a along the new dimension, the b branch stretches it by b.

Example 2 — walk Pascal row 4 from row 3

Row 3 of Pascal's triangle is 1, 3, 3, 1. The recurrence says row 4 is built by adding adjacent pairs (with implicit zeros at the edges):

\underbrace{1}_{0+1} \quad \underbrace{4}_{1+3} \quad \underbrace{6}_{3+3} \quad \underbrace{4}_{3+1} \quad \underbrace{1}_{1+0}

So row 4 is 1, 4, 6, 4, 1. Geometrically you have just extruded the cube into a 4D hypercube. The 4 comes from "1 + 3" — one a^3 corner extruded into a b-direction (becoming an a^3 b piece) plus three a^2 b slabs each extruded into an a-direction (each staying a^2 b, but now in 4D with one extra a, so a^3 b). The 6 is "3 + 3": three a^2 b slabs extruded into b plus three a b^2 slabs extruded into a, all becoming a^2 b^2 pieces.

Check by counting: 1 + 4 + 6 + 4 + 1 = 16 = 2^4. The 4D hypercube has exactly 16 pieces, and the row sums confirm it. So

(a+b)^4 = a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4

You did not multiply (a+b) four times; you walked Pascal's triangle one row.

Example 3 — why Pascal's recurrence is exactly the geometric merge

Take a generic piece in the n-cube of type "k b's" — say a^{n-k} b^k. There are \binom{n}{k} such pieces. Where did they come from in the (n-1)-cube, before you added the new axis?

There are exactly two possibilities. Either:

  • The new axis got an a. Then before extrusion, the piece had k b's already (the new a does not add a b). There were \binom{n-1}{k} such pieces.
  • The new axis got a b. Then before extrusion, the piece had only k - 1 b's (the new axis added the k-th b). There were \binom{n-1}{k-1} such pieces.

These two ancestor groups are disjoint (you cannot have both "new axis is a" and "new axis is b"), and they cover every type-k piece in the new row. So the count adds:

\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}

Pascal's recurrence is just disjoint cases for the new axis. Each entry sits below the two entries it inherits from. No magic — just bookkeeping for one extra dimension.

Why this matters

Once you see that squares and cubes are the same picture in different dimensions, three things become permanent:

  1. You stop seeing (a+b)^2 and (a+b)^3 as separate facts. They are the same dissection, stepped one dimension apart. Same goes for (a-b)^2 and (a-b)^3 — both are inclusion–exclusion, one in 2D and one in 3D.
  2. Pascal's triangle stops feeling like a memorisation table. It is a piece-counter for hypercubes. Row n tells you what an n-cube looks like after the standard slice.
  3. The binomial theorem becomes obvious. (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k is just "slice the n-hypercube of side a + b and count pieces by type". You can prove it for n = 2 with a square, for n = 3 with a cube, and the same logic continues forever.

That is the bridge. One extra axis, multiplied through. Pascal does the rest.

References

  1. Algebraic identities — the parent article.
  2. Geometric proof of (a+b)² — the 2D dissection.
  3. Geometric proof of (a+b)³ — the 3D dissection this article bridges to.
  4. Wikipedia: Pascal's triangle — the recurrence and its many faces.
  5. Wikipedia: Binomial theorem — the general n-dimensional statement.
  6. Cut the Knot: Sums of binomial coefficients — combinatorial identities tied to Pascal's triangle.