# How do you find three cube roots of -1?

Feb 22, 2017

Three cube roots of $1$ are $\left\{- 1 , \frac{1}{2} - \frac{\sqrt{3}}{2} i , \frac{1}{2} + \frac{\sqrt{3}}{2} i\right\}$

#### Explanation:

Let $x$ be the cube root of $- 1$, then we have $x \cdot 3 = - 1$

or ${x}^{3} + 1 = 0$

$\Leftrightarrow {x}^{3} + {x}^{2} - {x}^{2} - x + x + 1 = 0$

or ${x}^{2} \left(x + 1\right) - x \left(x + 1\right) + 1 \left(x + 1\right) = 0$

or $\left(x + 1\right) \left({x}^{2} - x + 1\right) = 0$

Hence either $x + 1 = 0$ i.e. $x = - 1$, or ${x}^{2} - x + 1 = 0$.

So one root is $x = - 1$ and for other two roots of ${x}^{2} - x + 1 = 0$, we proceed as follows:

${x}^{2} - x + 1 = 0 \Leftrightarrow {x}^{2} - 2 \times x \times \left(\frac{1}{2}\right) + {\left(\frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} + 1 = 0$

or ${\left(x - \frac{1}{2}\right)}^{2} + \frac{3}{4} = 0$

i.e. ${\left(x - \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{3}}{2} i\right)}^{2} = 0$

i.e. $\left(x - \frac{1}{2} + \frac{\sqrt{3}}{2} i\right) \left(x - \frac{1}{2} - \frac{\sqrt{3}}{2} i\right) = 0$

i.e. $x = \frac{1}{2} - \frac{\sqrt{3}}{2} i$ or $x = \frac{1}{2} + \frac{\sqrt{3}}{2} i$

Hence, three cube roots of $1$ are $\left\{- 1 , \frac{1}{2} - \frac{\sqrt{3}}{2} i , \frac{1}{2} + \frac{\sqrt{3}}{2} i\right\}$