How do you solve x^3-1=0?

1 Answer
Nov 30, 2016

The solutions are $S = \left\{1 , - \frac{1}{2} + i \frac{\sqrt{3}}{2} , - \frac{1}{2} - i \frac{\sqrt{3}}{2}\right\}$

Explanation:

We use $\left({a}^{3} - {b}^{3}\right) = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

Therefore.

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

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

$x = 1$

For the equation, ${x}^{2} + x + 1 = 0$

We calculate the discriminant,

$\Delta = {b}^{2} - 4 a c = 1 - 4 \cdot 1 \cdot 1 = - 3$

As, $\Delta > 0$, we do not a solution in $\mathbb{R}$ but in $\mathbb{C}$

$x = \frac{- 1 \pm \sqrt{\Delta}}{2} = \frac{- 1 \pm i \sqrt{3}}{2}$

${x}_{1} = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$ and ${x}_{2} = - \frac{1}{2} - i \frac{\sqrt{3}}{2}$