# How do you find all the real and complex roots of z^3 +1 = 0?

Jan 15, 2016

Factor then solve the remaining quadratic using the quadratic formula to find:

$z = - 1$ or $z = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

#### Explanation:

$0 = {z}^{3} + 1 = \left(z + 1\right) \left({z}^{2} - z + 1\right)$

Solve ${z}^{2} - z + 1 = 0$ using the quadratic formula with $a = 1$, $b = - 1$ and $c = 1$:

$z = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{1 \pm \sqrt{- 3}}{2} = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

Hence ${z}^{3} + 1 = 0$ has roots $z = - 1$, $z = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$