Question

How do I find the complex cube root of 8?

Answer 1

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

Answer 2

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

Answer 3

@Satellite73

Answer 4

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

Answer 5

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

Answer 6

@kerstie hello

Answer 7

ok thanks a lot, you helped me greatly

Answer 8

yw