(a + b)³ = a³ + 3a²b + 3ab² + b³

Quick Scoop

When people say “a plus b whole cube” , they mean the expression (a+b)3(a+b)³(a+b)3, i.e., the cube of a binomial. This identity lets you expand (a+b)3(a+b)³(a+b)3 without multiplying (a+b)(a+b)(a+b) three separate times.

The Core Formula

  • (a+b)3=(a+b)×(a+b)×(a+b)(a+b)³=(a+b)\times (a+b)\times (a+b)(a+b)3=(a+b)×(a+b)×(a+b)
  • Expanded form:
    (a+b)3=a3+3a2b+3ab2+b3(a+b)³=a³+3a²b+3ab²+b³(a+b)3=a3+3a2b+3ab2+b3

Here:

  • a3a³a3: cube of the first term.
  • b3b³b3: cube of the second term.
  • 3a2b3a²b3a2b and 3ab23ab²3ab2: mixed terms containing both aaa and bbb with coefficient 3.

How it’s Derived (Mini Walkthrough)

You can expand step by step:

  1. Start with three factors:
    (a+b)×(a+b)×(a+b)(a+b)\times (a+b)\times (a+b)(a+b)×(a+b)×(a+b).
  1. First multiply two brackets:
    (a+b)(a+b)=a2+2ab+b2.(a+b)(a+b)=a²+2ab+b².(a+b)(a+b)=a2+2ab+b2.
  1. Then multiply the result by (a+b)(a+b)(a+b):
    (a2+2ab+b2)(a+b)=a3+3a2b+3ab2+b3.(a²+2ab+b²)(a+b)=a³+3a²b+3ab²+b³.(a2+2ab+b2)(a+b)=a3+3a2b+3ab2+b3.

That’s why the shortcut identity works every time.

Fast Memory Trick

A simple way to remember the pattern:

  • First term: a3a³a3
  • Last term: b3b³b3
  • Middle terms follow decreasing powers of aaa and increasing powers of bbb, with binomial coefficients 3 and 3:
    • 3a2b3a²b3a2b
    • 3ab23ab²3ab2

So the power of aaa goes 3,2,1,03,2,1,03,2,1,0 while the power of bbb goes 0,1,2,30,1,2,30,1,2,3, and the coefficients are 1,3,3,11,3,3,11,3,3,1.

Tiny Example

Let a=2a=2a=2 and b=1b=1b=1:

  • Using the identity:
    (2+1)3=23+3⋅22⋅1+3⋅2⋅12+13(2+1)³=2³+3\cdot 2²\cdot 1+3\cdot 2\cdot 1²+1³(2+1)3=23+3⋅22⋅1+3⋅2⋅12+13
    =8+12+6+1=27.=8+12+6+1=27.=8+12+6+1=27.
  • Direct check:
    (2+1)3=33=27.(2+1)³=3³=27.(2+1)3=33=27.

Both match, so the identity holds.

Extra: Related Cube Identity

A related identity that’s often taught alongside is:

a3+b3=(a+b)(a2−ab+b2)a³+b³=(a+b)(a²-ab+b²)a3+b3=(a+b)(a2−ab+b2)

It can be derived starting from the plus b whole cube identity (a+b)3=a3+b3+3ab(a+b)(a+b)³=a³+b³+3ab(a+b)(a+b)3=a3+b3+3ab(a+b) and rearranging terms.

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