The real cube root of 1 is $\sqrt[3]{1} = 1$. But 1 has two more cube roots hiding in the complex numbers.
Quick Answer:
Result: $\sqrt[3]{1} = 1$ (the real cube root)
All three roots of $x^3 = 1$: $1,\ \omega = \dfrac{-1 + i\sqrt{3}}{2},\ \omega^2 = \dfrac{-1 - i\sqrt{3}}{2}$
Notation: $\sqrt[3]{1}$ or $1^{1/3}$
Method shown: factoring $x^3 - 1 = 0$
Exact form: $1$ (real); the other two are complex
Quick Reference Table
Number | Real cube root | Perfect cube? |
|---|---|---|
$\sqrt[3]{1}$ | $1$ | Yes |
$\sqrt[3]{8}$ | $2$ | Yes |
$\sqrt[3]{27}$ | $3$ | Yes |
$\sqrt[3]{64}$ | $4$ | Yes |
$\sqrt[3]{125}$ | $5$ | Yes |
$\sqrt[3]{-1}$ | $-1$ | Yes |
Where the Cube Roots of 1 Appear
The three cube roots of unity are spaced evenly around a circle of radius 1 in the complex plane, $120^\circ$ apart. They show up in the discrete Fourier transform used in signal processing, in group theory as the simplest non-trivial cyclic group, and in solving cubic equations by Cardano's method. Anywhere a rotation by a third of a turn matters, $\omega$ is the number doing the rotating.
What Is the Cube Root of 1?
The cube root of a number is the value that, multiplied by itself three times, gives that number. Since $1 \times 1 \times 1 = 1$, the real cube root of 1 is 1.
But "the cube root" hides something. The equation $x^3 = 1$ is a cubic, and a cubic always has three roots. One is the real number 1; the other two are complex. Together they are called the cube roots of unity, where unity just means the number 1. These build on the ideas in exponents and radicals.
How to Find the Three Cube Roots of 1 (Methods)
Method 1: Factor $x^3 - 1 = 0$
Start from the definition. Any cube root of 1 satisfies:
$$x^3 = 1$$ $$x^3 - 1 = 0$$
Factor the difference of cubes.
$$(x - 1)(x^2 + x + 1) = 0$$
The first factor gives the real root.
$$x - 1 = 0$$ $$x = 1$$
The second factor is a quadratic. Solve $x^2 + x + 1 = 0$ with the quadratic formula, where $a = 1$, $b = 1$, $c = 1$.
$$x = \frac{-1 \pm \sqrt{1 - 4}}{2}$$ $$x = \frac{-1 \pm \sqrt{-3}}{2}$$ $$x = \frac{-1 \pm i\sqrt{3}}{2}$$
Final answer: the three cube roots of 1 are
$$1, \quad \omega = \frac{-1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{-1 - i\sqrt{3}}{2}$$
Method 2: Use the two key properties
The complex roots satisfy two relationships worth knowing on sight.
The sum of all three roots is zero.
$$1 + \omega + \omega^2 = 0$$
The product of the three roots is 1.
$$1 \cdot \omega \cdot \omega^2 = \omega^3 = 1$$
These follow because one complex root is the square of the other: squaring $\omega$ produces $\omega^2$, and cubing either returns to 1.
Common Mistakes With Cube Root of 1
Mistake 1: Saying 1 has only one cube root
Where it slips in: when a student meets $\sqrt[3]{1}$ before complex numbers. Don't do this: stopping at $\sqrt[3]{1} = 1$ for the equation $x^3 = 1$. The correct way: the symbol $\sqrt[3]{1}$ means the principal (real) root, which is 1, but the equation $x^3 = 1$ has three solutions. Match the answer to the question.
Mistake 2: Dropping the cube-root index
Where it slips in: writing the radical in a hurry. Don't do this: writing $\sqrt{1}$ when you mean the cube root. The correct way: always show the index: $\sqrt[3]{1}$. The little 3 changes the meaning entirely.
Mistake 3: Forgetting $\omega^3 = 1$ when simplifying powers
Where it slips in: evaluating a high power like $\omega^7$. Don't do this: multiplying $\omega$ seven times. The correct way: reduce the exponent using $\omega^3 = 1$, so $\omega^7 = \omega^{6} \cdot \omega = (\omega^3)^2 \cdot \omega = \omega$.
To build cube-root fluency with a teacher, explore Bhanzu's algebra tutor or math classes online.
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