# How do you find all the real cube roots of 8/125?

Oct 19, 2015

#### Answer:

$\frac{8}{125}$ has one Real cube root $\frac{2}{5}$

#### Explanation:

$8 = {2}^{3}$ and $125 = {5}^{3}$

So

$\sqrt[3]{\frac{8}{125}} = \sqrt[3]{{2}^{3} / {125}^{3}} = \frac{\sqrt[3]{{2}^{3}}}{\sqrt[3]{{5}^{3}}} = \frac{2}{5}$

The other two roots are both Complex:

$\frac{2}{5} \omega = \frac{2}{5} \left(\frac{- 1 + i \sqrt{3}}{2}\right) = \frac{- 1 + i \sqrt{3}}{5}$

$\frac{2}{5} {\omega}^{2} = \frac{2}{5} \left(\frac{- 1 + i \sqrt{3}}{2}\right) = \frac{- 1 - i \sqrt{3}}{5}$

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