// maths › Cube Identities
Verify (a+b)³ − (a−b)³ = 6a²b + 2b³, where even-power terms cancel and odd ones double.
(a+b)³ - (a-b)³ = 6a² b + 2b³
A mind behind this: Pingala c. 300 BCE
Subtracting the two cubes does the opposite of adding them: now the even-in-b terms (a³ and 3ab²) cancel, and the odd-in-b terms double, leaving 6a²b + 2b³. So adding the cubes keeps the symmetric part, while subtracting keeps the antisymmetric part. Together, the sum and difference let you isolate either half of an expression - exactly the even/odd separation used throughout applied maths.
In (a+b)³ − (a−b)³, the a³ terms (both +a³) cancel, and the 3ab² terms (both +3ab²) cancel, because subtraction removes what is identical. But +3a²b minus −3a²b gives +6a²b, and +b³ minus −b³ gives +2b³ - the terms that had opposite signs now reinforce. So subtraction keeps exactly the terms that addition cancelled, and vice versa. The two identities are perfect complements.
In extracting the antisymmetric or 'odd' part of a relationship. In signal processing, the odd part of a signal is found by f(x) − f(−x) over 2, structurally identical to this difference of cubes. It also gives a quick way to compute differences of paired cubes mentally: 12³ − 8³ with a=10, b=2 is 6·100·2 + 2·8 = 1200 + 16 = 1216, instantly.
They are a matched pair: (a+b)³ + (a−b)³ = 2a³ + 6ab² captures the even-in-b part, and (a+b)³ − (a−b)³ = 6a²b + 2b³ captures the odd-in-b part. Add those two results and you recover 2(a+b)³; subtract them and you get 2(a−b)³. This is the algebraic skeleton of decomposing anything into symmetric and antisymmetric pieces and then reassembling it.
Yes - anywhere you need just the part of a change that behaves antisymmetrically. In physics, separating a force or field into symmetric and antisymmetric components simplifies analysis; in data, isolating the odd component of a paired difference highlights directional effects. The cube case is a clean, hand-checkable example of a technique that scales to functions and signals.