(a + b)³ and (a − b)³ Formulas with Examples

The (a + b)³ formula is (a + b)³ = a³ + 3a²b + 3ab² + b³, which can also be written as a³ + b³ + 3ab(a + b). Its partner, (a − b)³, equals a³ − 3a²b + 3ab² − b³.

These cube identities are a key part of Class 9 Polynomials. They also help with factorisation and quick calculations.

How (a + b)³ Is Derived

(a + b)³ = (a + b)² × (a + b). We already know (a + b)² = a² + 2ab + b² (see our (a + b)² formula guide). Now multiply:

(a² + 2ab + b²)(a + b) = a³ + a²b + 2a²b + 2ab² + ab² + b³ = a³ + 3a²b + 3ab² + b³

Take 3ab common from the middle two terms: 3a²b + 3ab² = 3ab(a + b)

So (a + b)³ = a³ + b³ + 3ab(a + b). This short form is very useful in exam questions.

The (a − b)³ Formula

Replace b with −b in the formula above: (a − b)³ = a³ − 3a²b + 3ab² − b³ = a³ − b³ − 3ab(a − b)

Sign pattern to remember: + − + − (the signs alternate, starting with plus).

Table of All Cube Identities

Name	Identity
Cube of a sum	(a + b)³ = a³ + 3a²b + 3ab² + b³
Short form	(a + b)³ = a³ + b³ + 3ab(a + b)
Cube of a difference	(a − b)³ = a³ − 3a²b + 3ab² − b³
Short form	(a − b)³ = a³ − b³ − 3ab(a − b)
Sum of cubes	a³ + b³ = (a + b)(a² − ab + b²)
Difference of cubes	a³ − b³ = (a − b)(a² + ab + b²)
Three-term identity	a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
Special case	If a + b + c = 0, then a³ + b³ + c³ = 3abc

Memory trick for a³ ± b³: the first bracket copies the sign of the question. The middle term of the second bracket takes the opposite sign, and the last term is always +.

Solved Examples

Example 1: Expand (x + 2)³
= x³ + 3(x²)(2) + 3(x)(2²) + 2³ = x³ + 6x² + 12x + 8

Example 2: Expand (2x − 3)³
= (2x)³ − 3(2x)²(3) + 3(2x)(3)² − 3³ = 8x³ − 36x² + 54x − 27

Example 3: Find 11³ without a calculator
11 = 10 + 1 → 1000 + 300 + 30 + 1 = 1331

Example 4: Find 99³
99 = 100 − 1 → 1000000 − 30000 + 300 − 1 = 970299

Example 5: If a + b = 5 and ab = 6, find a³ + b³
a³ + b³ = (a + b)³ − 3ab(a + b) = 125 − 90 = 35
Check: 2³ + 3³ = 8 + 27 = 35 ✓

Example 6: Factorise 8x³ + 27
8x³ + 27 = (2x)³ + 3³ = (2x + 3)(4x² − 6x + 9)

Example 7: Factorise x³ − 64
x³ − 4³ = (x − 4)(x² + 4x + 16)

Example 8: Find (−12)³ + 7³ + 5³ without cubing
−12 + 7 + 5 = 0, so the sum equals 3abc = 3 × (−12) × 7 × 5 = −1260

Common Mistakes to Avoid

• Writing (a + b)³ = a³ + b³. The middle terms 3a²b + 3ab² are missing.
• Forgetting to cube the coefficient: (2x)³ = 8x³, not 2x³.
• Mixing up the signs in a³ − b³. The second bracket is (a² + ab + b²).
• Using (a − b)³ with the signs + + + +. The signs must alternate: + − + −.

Practice Questions

1. Expand (y + 3)³
2. Expand (x − 1)³
3. Factorise 27a³ − 1
4. If a − b = 4 and ab = 5, find a³ − b³

Answers: 1. y³ + 9y² + 27y + 27   2. x³ − 3x² + 3x − 1   3. (3a − 1)(9a² + 3a + 1)   4. 124