How do you get the complex cube root of 8?

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Answer 1 · 2016-01-03T21:05:01

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 the quadratic formula:

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

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

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

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