# Cube Root of 27 — Simplified Form, Methods, and Proof

TL;DR

The cube root of 27 is exactly 3, because 3×3×3=27. 27 is a perfect cube, so the cube root is a clean integer — no decimal approximation needed. This article covers the value, the prime-factorisation method, where the cube root shows up, and the slips students make most often.

## The Answer At A Glance

**Result:** 3

**Notation:** \(27^{1/3}\) or \(3\sqrt[3]{27}

**Method shown:** Prime factorisation, with cross-checks using direct verification and consecutive-cube bracketing.

**Exact form:** 3 — an integer.

**Type:** 27 is a **perfect cube** because \(3^3 = 27.

## Quick Reference Table — Small Perfect Cubes

| n   | n³       | ∛n |
|-----|----------|----|
| 1   | 1        | 1  |
| 2   | 8        | 2  |
| **3** | **27** | **3**  |
| 4   | 64       | 4  |
| 5   | 125      | 5  |
| 6   | 216      | 6  |
| 7   | 343      | 7  |
| 8   | 512      | 8  |
| 9   | 729      | 9  |
| 10  | 1000     | 10 |

27=3³ is the third entry in the perfect-cube sequence, after 1 and 8.

## What "Cube Root of 27" Means

The cube root of a number \(n\) is the value \(x\) such that \(x^3=n\). For \(\sqrt[3]{27}\), the \(x\) with \(x \cdot x \cdot x = 27\).

Verification: \(3\cdot 3 = 9\). \(9\cdot 3 = 27\). So \(3^3 = 27,\) confirming \(\sqrt[3]{27}=3\).

Unlike a square root, the cube root of a negative number is real: \(\sqrt[3]{-27} = -3\).

## How to Find ∛27 — Three Methods

### Method 1 — Prime factorisation

Factor 27 into primes:

\(27=3\cdot 9=3\cdot 3\cdot 3=3^3\)

Group the prime factors in triples:

\(\sqrt[3]{27} = \sqrt[3]{3^3} = 3\)

### Method 2 — Direct verification

\(3^3 = 3 \cdot 3 \cdot 3 = 27✓\)

### Method 3 — Bracketing by consecutive cubes

\(2^3=8,3^3=27,4^3=64\)

\(27=3^3\) exactly. So \(\sqrt[3]{27} = 3\).

## Is ∛27 Rational or Irrational?

\(\sqrt[3]{27}\) is **rational** — an integer. 27 is a perfect cube, and the cube root of a perfect cube is always a whole number.

## A Proof Using The Algebraic Identity

For a quick algebraic confirmation, use the identity \(a^3−b^3=(a−b)(a^2+ab+b^2)\).

Setting \(a^3=27\) and \(b=3\):

\(27−27=(a−3)(a^2+3a+9)\).

The left side is zero. For the right side to be zero, \(a−3=0\) confirming \(\sqrt[3]{27}=3\).

## Where ∛27 Shows Up

\(\sqrt[3]{27}\) comes up whenever you find cubic relationships.

- **Cube volumes.** A cube with volume 27 cubic units has a side length of 3 units.
- **Scaling solids.** Tripling the linear dimensions of a 3D shape scales the volume by 27. 
- **Polynomial roots.** The equation \(x^3=27\) has the real solution x=3.
- **Sum of cubes.** 27+1=28 is not a perfect cube.

## Tripping Points To Avoid On Cube-Root Problems

### **Mistake 1: Confusing the cube root with the square root.**

- **Where it slips:** Reading \(\sqrt[3]{27}\) but reaching for the square-root key.
- **The correct way:** \(\sqrt[3]{27} = 3\) is integer, while \(\sqrt{27} \approx 5.196\).

### **Mistake 2: Squaring instead of cubing during verification.**

- **Where it slips:** Verifying \(3\) by computing \(3^2\) instead of \(3^3\).
- **The correct way:** Verify by cubing: \(3^3 = 27✓\).

### **Mistake 3: Forgetting the cube root of a negative is well-defined.**

- **Where it slips:** Refusing to compute \(-\sqrt[3]{-27}\) on the grounds that "roots of negatives are not real."
- **The correct way:** \(-\sqrt[3]{-27} = -3\).

## Conclusion

- The **cube root of 27** is exactly 3.
- 27 is a perfect cube and its cube root is a clean integer.
- Three methods confirm the value: prime factorisation, direct verification, and consecutive cubes.
- The cube root of a negative number is well-defined: \(-\sqrt[3]{-27} = -3\).

## Frequently Asked Questions

**What is the cube root of 27?**
\(\sqrt[3]{27} = 3\).

**Is 27 a perfect cube?**
Yes, because \(27=3^3\).

**How do you find the cube root of 27?**
Prime-factorise: \(27=3^3\), and the cube root is 3.

**What is \(-\sqrt[3]{-27}\)?**
\(-\sqrt[3]{-27} = -3\).

**Is the cube root of 27 rational or irrational?**
Rational — it is an integer.