# How do you solve x^3+1/x^3 = 0 ?

Aug 17, 2016

${x}_{1 , 2} = \frac{\sqrt{3}}{2} \pm \frac{1}{2} i$

${x}_{3 , 4} = \pm i$

${x}_{5 , 6} = - \frac{\sqrt{3}}{2} \pm \frac{1}{2} i$

#### Explanation:

Multiply through by ${x}^{3}$ to find:

${x}^{6} + 1 = 0$

For any Real value of $x$ we have ${x}^{6} \ge 0$, hence ${x}^{6} + 1 \ne 0$

$\textcolor{w h i t e}{}$
Complex solutions

From de Moivre we have:

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta$

Hence Complex solutions:

$x = \cos \left(\frac{\left(2 k + 1\right) \pi}{6}\right) + i \sin \left(\frac{\left(2 k + 1\right) \pi}{6}\right)$

That is:

${x}_{1 , 2} = \frac{\sqrt{3}}{2} \pm \frac{1}{2} i$

${x}_{3 , 4} = \pm i$

${x}_{5 , 6} = - \frac{\sqrt{3}}{2} \pm \frac{1}{2} i$

These six roots form the vertices of a regular hexagon in the Complex plane.

graph{((x-sqrt(3)/2)^2+(y-1/2)^2-0.002)((x-sqrt(3)/2)^2+(y+1/2)^2-0.002)((x+sqrt(3)/2)^2+(y-1/2)^2-0.002)((x+sqrt(3)/2)^2+(y+1/2)^2-0.002)(x^2+(y-1)^2-0.002)(x^2+(y+1)^2-0.002) = 0 [-2.5, 2.5, -1.25, 1.25]}