How do I find the complex cube root of 8?

Question

anonymous · Answer

method one
solve 
$$x^3=8$$
$$x^3-8=0$$
$$(x-2)(x^2+2x+4)=0$$
quadratic formula will give the solutions to 
$$x^2+2x+4=0$$

anonymous · Answer

I see and so 2 is one root and the other 2 complex numbers will be the other 2 roots?

anonymous · Answer

@Satellite73

anonymous · Answer

method two
 
solve $x^3=1$ and multiply by two 
since one solution is 1, divide the unit circle in two three equal parts
one is at 1, the other two are $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and its conjugate $-\frac{1}{2}-\frac{\sqrt{3}}{2}i$
multiply  these numbers by 2

anonymous · Answer

to answer your question, yes, the solution to the quadratic will give you the two complex roots

anonymous · Answer

@kerstie hello

anonymous · Answer

ok thanks a lot, you helped me greatly

anonymous · Answer

yw