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

[Math Formula](/content/tag/math-formula/index.html)

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.**  
> - **Type:** algebraic identity (true for all real a, b).  
> - **Reads as:** "sum of cubes equals sum-times-(square-minus-product-plus-square)."  
> - **Grade introduced:** [CCSS-M HSA-SSE.A.2](https://www.thecorestandards.org/Math/Content/HSA/SSE/) (seeing structure in expressions); [NCERT Class 9 Chapter 2 — Polynomials](https://ncert.nic.in/textbook.php) Identity VII.  
> - **Related identity:** a³ − b³ = (a − b)(a² + ab + b²).
> - **Three-variable version:** a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

## 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.

- **Volume problems.** Adding the volumes of two cubes of side lengths a and b gives a³ + b³. Factoring expresses total volume as the side-sum times an irreducible quadratic.
- **Polynomial root-finding.** Every cubic of the form x³ + k = 0 uses this identity to spot one real root.
- **Number theory.** Fermat's Last Theorem for n=3 was proved by Euler in 1770, and the proof leans on the algebraic structure of the cube identity.
- **Physics — heat capacity at low temperatures.** Expressions factor through identities like this one when computing differences of cubic terms.
- **Engineering — pipe-volume calculations.** Total cubic-foot volume often factors via the identity.

## 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
- The **a cube plus b cube formula** is a³ + b³ = (a + b)(a² − ab + b²).
- The companion identity is a³ − b³ = (a − b)(a² + ab + b²).
- SOAP prevents the most common factoring error.
- The trinomial factor a² − ab + b² is irreducible over the real numbers.

## Five Minutes of Practice — Three Problems
1. Factorise x³ + 64.
2. Factorise 27a³ + 8b³.
3. Factorise x⁶ + 1 over the reals.