# How do I find the three cube roots of unity?

Apr 1, 2015

Although you could use de Moivre's Threorem, or another version of the geometry of the complex plane, it's not needed if you know and apply some algebra:

The cube roots on unity (1) are the complex solutions to
${x}^{3} = 1$

${x}^{3} - 1 = 0$

Now factor

Method 1: use ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$ to get
${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right)$

Method 2: Use the (obvious?) fact that $1$ is a zero of ${x}^{3} - 1$ to see that $\left(x - 1\right)$ must be a factor.
Then do the division (long or synthetic) (or use trial and error) to get ${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right)$

So we need to solve:
${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right) = 0$.

Use your favorite applicable method to solve: $\left({x}^{2} + x + 1\right) = 0$