Cube Roots of Unity Problems
On this page
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.