# What is the cube root of -0.000125 ?

Mar 3, 2017

$\sqrt[3]{- 0.000125} = - 0.05$

#### Explanation:

$- 0.000125 = - 125 \cdot {10}^{- 6} = - {5}^{3} \cdot {10}^{- 6}$

This has one Real cube root and two complex ones.

$\sqrt[3]{- {5}^{3} \cdot {10}^{- 6}} = - 5 \cdot {10}^{- 2} = - 0.05$

The two complex cube roots are:

$- 0.05 \omega \text{ }$ and $\text{ } - 0.05 {\omega}^{2}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive complex cube root of $1$.

$\omega$ and ${\omega}^{2}$ are the zeros of:

$\frac{{x}^{3} - 1}{x - 1} = {x}^{2} + x + 1$