# Cube Roots of Unity Problems

> Cube roots of unity for JEE Mathematics: ω³ = 1, 1 + ω + ω² = 0, power reduction modulo 3, factorisations and classic evaluation problems.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/cube-roots-unity-problems
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Cube Roots of Unity Problems", PrepElephant, https://prepelephant.com/topics/jee/mathematics/cube-roots-unity-problems

## Direct answer

Cube roots of unity are 1, ω, ω², where ω = −1/2 + i√3/2; the two facts that unlock every problem are ω³ = 1 and 1 + ω + ω² = 0, with ω and ω² mutual conjugates forming an equilateral triangle on the Argand plane. Expressions collapse by reducing powers modulo 3 — ω^(3k) = 1, ω^(3k + 1) = ω, ω^(3k + 2) = ω² — and by rewriting 1 + ω as −ω² and 1 + ω² as −ω. The algebra ties: x³ − 1 = (x − 1)(x − ω)(x − ω²), x² + x + 1 = (x − ω)(x − ω²), and x^(3k) − 1 is divisible by x² + x + 1.

## What you must remember

- **The pair of identities:** ω³ = 1 and 1 + ω + ω² = 0 — between them they kill every expression in ω.
- **Rewrites worth reflexes:** 1 + ω = −ω², 1 + ω² = −ω, ω − ω² = i√3, ω × ω² = 1, |ω| = |ω²| = 1.
- **Conjugate symmetry:** ω² is the conjugate of ω, so ω + ω² = −1 and any expression symmetric under swapping ω with ω² is real.
- **Power reduction:** divide the exponent by 3 and keep the remainder — ω^2027 = ω^(3 × 675 + 2) = ω².
- **Factorisations:** x³ − y³ = (x − y)(x − ωy)(x − ω²y), and x² + xy + y² = (x − ωy)(x − ω²y) for the same reason.
- **Divisibility rule:** x² + x + 1 divides x^n − 1 exactly when n is a multiple of 3 — remainder computations reduce to x³ = 1.
- **Geometry:** the three cube roots sit at 120° intervals on the unit circle, an equilateral triangle whose centroid is the origin.

## Collapsing powers of ω

Evaluate (1 + ω − ω²)³ − (1 − ω + ω²)³. Rewrite each bracket using the sum identity: 1 + ω − ω² = (1 + ω + ω²) − 2ω² = −2ω², and 1 − ω + ω² = −2ω. The cubes follow: (−2ω²)³ = −8ω⁶ = −8, since ω⁶ = (ω³)² = 1, and (−2ω)³ = −8ω³ = −8. The difference is −8 − (−8) = 0. Now the same reflex one power up: (1 + ω − ω²)⁷ = (−2ω²)⁷ = −128 ω¹⁴, and 14 = 3 × 4 + 2 leaves ω², so the value is −128ω² — a complex number the options quote in full a + ib form as 64 + 64i√3 (multiply −128 by 1/2 − i√3/2? Check: −128ω² = −128(−1/2 − i√3/2) = 64 + 64i√3). The method never varies: force the brackets into single powers of ω through 1 + ω + ω² = 0, then let the exponent fall modulo 3.

## ω in factor problems

JEE Main asks the evaluations above as single-correct items, and the distractors exploit two habits: computing ω² as if it were a positive real (dropping the conjugate's negative real part) and reducing exponents modulo 2 instead of 3. The value −128ω² quoted in a + ib form catches everyone who never converted back from polar clothing. JEE Advanced prefers the polynomial face: find the remainder when x^2027 + x + 1? Reduce modulo x² + x + 1 by setting x³ = 1: 2027 = 3 × 675 + 2, so x^2027 ≡ x² ≡ −x − 1, making the expression (−x − 1) + x + 1 ≡ 0 — the polynomial is divisible, and the reasoning took one line of modulo arithmetic. The sum-geometry items appear too: the value of (a + bω + cω²)/(a − bω − cω²)? type questions resolve by multiplying numerator and denominator by conjugates, using ω·ω̄ = ω·ω² = 1. The discipline that holds all of it: write the two identities at the top of the rough sheet before touching any ω expression, and convert 1 + ω sightings into −ω² on sight, before arithmetic buries them.

## Frequently asked questions

### What are the two governing identities of cube roots of unity?

ω³ = 1 and 1 + ω + ω² = 0 — together they reduce every polynomial expression in ω to a + bω form.

### How do you simplify ω raised to a large power?

Reduce the exponent modulo 3: ω^(3k) = 1, ω^(3k + 1) = ω, ω^(3k + 2) = ω².

### What is the value of 1 + ω − ω²?

−2ω², obtained by rewriting the expression as (1 + ω + ω²) − 2ω² — and its cube is −8.

### How does x² + x + 1 relate to cube roots of unity?

It factors as (x − ω)(x − ω²), so it divides x^n − 1 exactly when n is a multiple of 3.

### Where do the cube roots of unity sit on the Argand plane?

At the vertices of an equilateral triangle inscribed in the unit circle, at angles 0, 120° and 240° — ω and ω² as conjugates.