a³ + b³ Formula — Sum of Cubes, Proof, Examples

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a³ + b³ Formula — Sum of Cubes, Proof, Examples

Math Formula

TL;DR
The a cube plus b cube formula — a³ + b³ = (a + b)(a² − ab + b²) — factors the sum of two cubes into a binomial times a trinomial. This article gives the formula, its algebraic proof, three worked examples at three difficulty tiers, a side-by-side with a³ − b³, the common factoring slips, and where the identity shows up in engineering and physics.

A Formula That Turns "Cubes Don't Factor" Into "Yes, They Do"

Most students meet a² − b² = (a − b)(a + b) in Class 8 and assume that's the whole story for factoring powers. Then they meet a sum of cubes — x³ + 8, say — and freeze. The expression looks unfactorable. It isn't.

The a cube plus b cube formula says:

a³ + b³ = (a + b)(a² − ab + b²).

One sum of cubes becomes one linear factor times one quadratic factor. Every sum of cubes — numerical or algebraic — yields to this identity.

The Formula

For real numbers a and b:

[ a³ + b³ = (a + b)(a² - ab + b²) ]

The companion identity for the difference of cubes is:

[ a³ − b³ = (a − b)(a² + ab + b²) ]

Notice the sign pattern. The binomial factor matches the sign on the left side (plus for sum, minus for difference). The middle term in the trinomial does the opposite (minus for sum, plus for difference). This is the single sign rule that catches most students out.

Quick facts.

How the Formula Is Derived — A Two-Line Proof

The cleanest proof multiplies out the right-hand side and watches the cross-terms cancel.

[(a+b)(a²−ab+b²) = a⋅a² + a⋅(−ab) + a⋅b² + b⋅a² + b⋅(−ab) + b⋅b² = a³+b³.]

The proof is the identity. Once a student has seen the cancellation, the formula stops feeling memorised and starts feeling inevitable.

Three Worked Examples — Quick, Standard, Stretch

Quick. Factorise x³ + 27.

Recognise 27 = 3³, so the expression is x³ + 3³. Apply the formula with a = x and b = 3.

x³ + 27 = (x + 3)(x² − 3x + 9).

Final answer: (x + 3)(x² − 3x + 9).

Standard. Factorise 8x³ + 125.

A student treats 8x³ as (8x)³ and writes 8x³ + 125 = (8x + 5)(64x² − 40x + 25). The flaw: 8x³ is (2x)³, not (8x)³. The rescue: 8x³ = (2x)³ and 125 = 5³. So:

8x³ + 125 = (2x + 5)(4x² − 10x + 25).

Check: (2x + 5)(4x² − 10x + 25) expanded gives 8x³ + 125.

Final answer: (2x + 5)(4x² − 10x + 25).

Stretch. Factorise a⁶ + b⁶ over the reals.

Rewrite as a sum of cubes: a⁶ + b⁶ = (a²)³ + (b²)³. Apply the formula with a² in place of a and b² in place of b:

a⁶ + b⁶ = (a² + b²)(a⁴ − a²b² + b⁴).

Final answer: (a² + b²)(a⁴ − a²b² + b⁴). Neither factor factors further over the reals.

a³ + b³ vs a³ − b³ — Side by Side

Identity Binomial factor Trinomial factor (middle term)
a³ + b³ (sum of cubes) (a + b) — plus a² − ab + b² — minus
a³ − b³ (difference of cubes) (a − b) — minus a² + ab + b² — plus

The mnemonic Bhanzu trainers use on the McKinney center whiteboard: "SOAP"Same sign in the binomial, Opposite sign on the middle term, Always Positive on the squared terms. Same-Opposite-Always-Positive.

Where the a³ + b³ Formula Shows Up

The identity appears wherever cubic structure does.

Tripping Points to Avoid

Mistake 1: Flipping the middle-term sign.

Incorrect: A student factors a³ + b³ as (a + b)(a² + ab + b²).
Correct: Use SOAP. Sum identity → trinomial middle term is minus.

Mistake 2: Cube-rooting the coefficient instead of the term.

Incorrect: Factoring 27x³ + 64 as a = 27x and b = 64.
Correct: Find a and b such that a³ = first term and b³ = second.

Mistake 3: Forgetting that the trinomial factor is usually irreducible.

Don't do this: Assume every quadratic breaks down.
The correct way: Note that the quadratic factor is irreducible over the reals.

Conclusion

Five Minutes of Practice — Three Problems

  1. Factorise x³ + 64.
  2. Factorise 27a³ + 8b³.
  3. Factorise x⁶ + 1 over the reals.