Introduction

The cubic equation x³ – 1 = 0 is a classic problem in complex numbers and cube roots of unity, frequently asked in competitive exams.
Understanding how to factor this equation and interpret its complex roots builds a strong conceptual grip on roots of unity and Argand‑plane geometry.
This article explains how to find all roots of x³ – 1 = 0 and evaluate each multiple‑choice option with clear, exam‑oriented reasoning.

Detailed Solution of x³ – 1 = 0

Start by factoring x³ – 1 as a difference of cubes:

x³ – 1 = (x – 1)(x² + x + 1) = 0.

The first factor gives the real root x = 1.

For the quadratic x² + x + 1 = 0, use the quadratic formula:

x = [-1 ± √(1 – 4)] / 2 = [-1 ± √(-3)] / 2 = -1/2 ± (√3/2)i.

Thus, the three roots are:

• 1
• -1/2 + (√3/2)i
• -1/2 – (√3/2)i

These two complex roots are the non‑real cube roots of unity.

MCQ Option‑wise Analysis

Option (A): -1/2 + (√3/2)i

The complex number -1/2 + (√3/2)i has modulus 1 and argument 2π/3.
Cubing a complex number with modulus 1 multiplies its argument by 3, giving argument 2π, which corresponds to 1 on the complex plane.
Therefore, (-1/2 + (√3/2)i)³ = 1, so option (A) is a correct root of x³ – 1 = 0.

Option (B): i

For x = i, compute x³:

i² = -1 and i³ = i²·i = -1·i = -i.
Since -i ≠ 1, the value i does not satisfy x³ – 1 = 0, so option (B) is incorrect.

Option (C): -i

For x = -i, compute x³:

(-i)² = (-1)²·i² = -1 and (-i)³ = (-i)²·(-i) = (-1)·(-i) = i.
Because i ≠ 1, -i is not a root of x³ – 1 = 0, so option (C) is also incorrect.

Option (D): -1/2 – (√3/2)i

The complex number -1/2 – (√3/2)i has modulus 1 and argument 4π/3.
Cubing it multiplies the argument by 3, giving 4π, which again corresponds to 1 on the complex plane.
Thus, (-1/2 – (√3/2)i)³ = 1, so option (D) is also a correct root of x³ – 1 = 0.

Therefore, when one solution is 1, the other solutions in the MCQ are given by options (A) and (D).