How do you get the complex cube root of 8?

The cube roots of ##8## are ##2##, ##2omega## and ##2omega^2## where ##omega=-1/2+sqrt(3)/2 i## is the primitive Complex cube root of ##1##.

Here are the cube roots of ##8## plotted in the Complex plane on the circle of radius ##2##:

graph{(x^2+y^2-4)((x-2)^2+y^2-0.01)((x+1)^2+(y-sqrt(3))^2-0.01)((x+1)^2+(y+sqrt(3))^2-0.01) = 0 [-5, 5, -2.5, 2.5]}

They can be written as:

##2(cos(0)+i sin(0)) = 2##

##2(cos((2pi)/3) + i sin((2pi)/3)) = -1 + sqrt(3)i = 2omega##

##2(cos((4pi)/3) + i sin((4pi)/3)) = -1 - sqrt(3)i = 2omega^2##

One way of finding these cube roots of ##8## is to find all of the roots of ##x^3-8 = 0##.

##x^3-8 = (x-2)(x^2+2x+4)##

The quadratic factor can be solved using :

##x = (-b +-sqrt(b^2-4ac))/(2a)##

##= (-2+-sqrt(2^2-(4xx1xx4)))/(2*1)##

##=(-2+-sqrt(-12))/2##

##=-1+-sqrt(3)i##