Algebra
Intermediate

Sum of Cubes

Factors the sum of two cubes.

Formula

a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a+b)(a^2 - ab + b^2)

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Variables

a, bAny real numbers

Example

8 + 27 = 2³ + 3³ = (2+3)(4 − 6 + 9) = 5×7 = 35. ✓

How it works

This identity rewrites a sum of two perfect cubes as a product of two factors, which makes the expression easier to simplify, solve, or factor. It works because expanding (a+b)(a^2-ab+b^2) gives a^3 + a^2b - a^2b - ab^2 + ab^2 + b^3, and the middle terms cancel, leaving a^3 + b^3. In other words, it is a special factoring pattern that always holds for real numbers a and b.

When to use it

Use this when you see an expression like x^3 + y^3 and want to factor it completely. It is especially useful in algebraic simplification, solving polynomial equations, and reducing expressions in calculus or higher-level math.

Worked example, step by step

Factor 64 + 125. 64 + 125 = 4^3 + 5^3 Use the sum of cubes formula: a^3 + b^3 = (a+b)(a^2 - ab + b^2) So, (4+5)(4^2 - 4·5 + 5^2) = 9(16 - 20 + 25) = 9(21) = 189

Common mistakes to avoid

  • Forgetting the middle term is negative: a^2 - ab + b^2, not a^2 + ab + b^2
  • Using the formula for a difference of cubes instead of a sum of cubes
  • Not recognizing that numbers like 64 and 125 are perfect cubes before factoring

Did You Know?

The sum and difference of cubes factor neatly — a fact exploited constantly in calculus and cryptography.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026