Sum of Cubes
Factoring formula for a sum of two cubes.
Formula
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Example
8+27 = (2+3)(4-6+9) = 35
How it works
This identity rewrites a sum of two perfect cubes as a product of a binomial and a quadratic. It works because if you expand (a+b)(a^2-ab+b^2), the middle terms cancel: a^2b and -a^2b, as well as -ab^2 and ab^2, leaving only a^3+b^3. This makes it a useful factoring pattern whenever you notice an expression that is a cube plus another cube.
When to use it
Use this formula when you need to factor expressions like x^3+8 or 27y^3+z^3, especially in simplifying algebraic fractions or solving polynomial equations. It is also helpful for spotting hidden structure in problems that involve sums of cubic terms.
Worked example, step by step
Factor 64 + 125 64 = 4^3 and 125 = 5^3, so this is a^3 + b^3 with a = 4 and b = 5 Apply the formula: a^3 + b^3 = (a+b)(a^2 - ab + b^2) (4+5)(4^2 - 4·5 + 5^2) (9)(16 - 20 + 25) 9(21) = 189
Common mistakes to avoid
- Forgetting that the middle sign in the quadratic factor is negative
- Using the formula on a difference of cubes instead of a sum of cubes
- Not recognizing perfect cubes first, such as 27 = 3^3 or 216 = 6^3
Did You Know?
The sum-of-cubes factorisation was known to medieval Islamic mathematicians centuries before modern algebraic notation existed.
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