In short
(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 is literally an (a+b) \times (a+b) \times (a+b) cube. Make one cut along each of the three axes — at the points where a ends and b begins — and the cube falls apart into eight rectangular boxes: one a \times a \times a corner, three a \times a \times b slabs (one for each axis), three a \times b \times b slabs, and one b \times b \times b corner. The coefficients 1, 3, 3, 1 are not memorised — they are counted off the picture.
You already know the square version: (a+b)^2 is a square of side a+b chopped into four tiles by one horizontal and one vertical cut. The cube version is the same idea, just one dimension up. Take a cube of side a+b. Make three cuts — one parallel to each face — at the spots where each edge splits from a into b. The cube does not turn into a mess. It falls apart into exactly eight neat rectangular boxes, like a Rubik's cube whose layers do not match. And the volumes of those eight boxes, added up, are exactly a^3 + 3a^2b + 3ab^2 + b^3.
Once you can see those eight boxes in your head, the identity is no longer something you have to memorise. It is something you can read straight off a 3D picture.
Building the cube, one cut at a time
Start with a solid cube of side length a+b. Its volume is (a+b)^3 — that is the whole point of the identity, the thing we want to compute.
Now mark the splitting point on every edge. On each of the twelve edges, draw a tick at distance a from one corner; the remaining b runs to the opposite corner. Connect the ticks across each face, and then across the inside, with three flat cuts:
- one horizontal cut (parallel to the floor),
- one vertical cut from front to back,
- one vertical cut from left to right.
Three perpendicular slices, three axes, one cut each. The cube falls into 2 \times 2 \times 2 = 8 smaller boxes. Why eight? Each axis is split into two pieces (a and b), and the three axis splits are independent. The number of resulting boxes is 2 \times 2 \times 2 = 8 — one box per choice of "a or b" along each axis.
Counting the eight pieces
Here is the bookkeeping that turns the picture into the identity. Every one of the eight boxes has three side lengths, each of which is either a or b. So you can name a box by a three-letter string like aab or bab — one letter per axis. There are 2^3 = 8 such strings, and they group naturally by how many a's they contain:
- Three a's (aaa): exactly one box, with volume a \cdot a \cdot a = a^3. Why one? There is only one way to choose "a" on every axis. That single box is the deep corner where all three a-pieces meet.
- Two a's, one b (aab, aba, baa): three boxes, each with two edges of length a and one of length b. Each volume is a^2 b. Why three? You are choosing which one of the three axes carries the b. Three axes, three choices, three slabs — one stuck to each face of the a^3 corner.
- One a, two b's (abb, bab, bba): three boxes, each with one a and two b edges. Each volume is a b^2. Why three again? Now you choose which of the three axes carries the a. Same Pascal-style symmetry, mirrored.
- Three b's (bbb): one box of volume b^3, the small corner diagonally opposite to the a^3 one.
Add the volumes:
That is the identity. The coefficients 1, 3, 3, 1 are just the number of pieces of each type. Why does the total volume equal (a+b)^3? Volume is invariant under dissection — when you cut a solid into non-overlapping pieces, the sum of the piece volumes equals the volume of the original. The eight pieces tile the cube perfectly with no gaps and no overlap, so their volumes must sum to (a+b)^3.
Worked examples
Example 1 — checking with a = 2, b = 1
A cube of side a + b = 3 has volume 27. The identity says
You can see all four contributions in a 3 \times 3 \times 3 arrangement of unit cubes (think of a small Rubik's cube): one 2 \times 2 \times 2 block in one corner (8 unit cubes), three flat 2 \times 2 \times 1 slabs glued to its three exposed faces (4 unit cubes each, so 12 total), three 2 \times 1 \times 1 rods filling the remaining edges (2 unit cubes each, so 6 total), and one lonely 1 \times 1 \times 1 cube at the far corner. 8 + 12 + 6 + 1 = 27. The Rubik's-cube picture is the identity, with side lengths a = 2 and b = 1.
Example 2 — why the coefficients are 1, 3, 3, 1 (Pascal's third row)
Look at the row of binomial coefficients
at the top of Pascal's triangle's third row. This is not a coincidence; it is the same counting you just did. The coefficient of a^{3-k} b^k in (a+b)^3 is the number of ways to choose which k of the three axes get a b — that is, \binom{3}{k}.
- \binom{3}{0} = 1 way to give zero axes a b → one a^3 corner.
- \binom{3}{1} = 3 ways to pick which one axis is a b → three a^2 b slabs.
- \binom{3}{2} = 3 ways to pick which two axes are b's → three a b^2 slabs.
- \binom{3}{3} = 1 way to make all three b → one b^3 corner.
The 1, 3, 3, 1 is a count of pieces in the cube. Pascal's triangle and the dissection of an n-cube are the same fact. Why the symmetry \binom{3}{k} = \binom{3}{3-k}? Swapping every "a" for "b" and vice versa turns the cube inside out: the a^3 corner becomes a b^3 corner, the a^2 b slabs become a b^2 slabs. The coefficients are mirror-symmetric because the picture is.
Example 3 — mental arithmetic: $11^3$ in five seconds
Write 11 = 10 + 1 and use the identity with a = 10, b = 1:
Notice the digits of the answer: 1, 3, 3, 1. The same Pascal row that counts the pieces in the cube literally appears as the digits of 11^3 — because 11 = 10 + 1 and powers of 10 shift digits left without overlap. The same trick gives 11^4 = 14641 (1, 4, 6, 4, 1 from Pascal's fourth row), and 11^5 = 161051 — which is almost 1, 5, 10, 10, 5, 1 except the two 10s carry over. Cricket scoring trick: an over has 6 legal balls, so 6 + 1 = 7 deliveries with one no-ball; 7^3 = (6+1)^3 = 216 + 108 + 18 + 1 = 343 in your head, no calculator.
The same shortcut works for 21^3 = (20+1)^3 = 8000 + 1200 + 60 + 1 = 9261 and 101^3 = (100+1)^3 = 1{,}000{,}000 + 30{,}000 + 300 + 1 = 1{,}030{,}301.
Why this matters
Once the cube picture is in your head, three things become permanent:
- You stop forgetting the middle coefficients. The two middle terms of (a+b)^3 both have coefficient 3, never 2 and never 4 — because there are exactly three axes in 3D space, so exactly three slabs of each mixed type. The "3" is geometric, not arbitrary.
- You can extend the trick. (a+b)^4 is a 4D hypercube cut into 2^4 = 16 pieces, with type counts 1, 4, 6, 4, 1 — the next row of Pascal's triangle, and also \binom{4}{0}, \binom{4}{1}, \dots, \binom{4}{4}. You cannot draw it, but the bookkeeping is identical: count the strings of as and bs of each composition.
- You see the binomial theorem coming. (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k is just "cut an n-dimensional hypercube of side a+b along all n axes; count the pieces of each type." That is the binomial theorem — and you have already proved it for n = 3 by drawing one cube.
The (a-b)^3 identity has its own picture too: start with the a^3 cube, slice off three slabs (one per face), notice you over-removed three rods, add them back, then notice you under-removed the corner, fix it. The result is (a-b)^3 = a^3 - 3a^2 b + 3 a b^2 - b^3, with the alternating signs explained exactly the same way the (a-b)^2 corner is — it is inclusion–exclusion in 3D.
References
- Algebraic identities — the parent article.
- Geometric proof of (a+b)² — the 2D version of the same dissection idea.
- Wikipedia: Binomial theorem — the general statement that (a+b)^n has coefficients given by Pascal's triangle.
- Wikipedia: Pascal's triangle — where 1, 3, 3, 1 lives, and why the symmetry is automatic.
- NCERT Class 9 Mathematics, Chapter 2: Polynomials — the Indian school treatment of (a+b)^3 and (a-b)^3.
- Cut the Knot: (a+b)³ visual proof — an interactive 3D dissection of the same cube.