SOLUTION: I'm 27 and not in school anymore. I didn't want you to think that I'm cheating on homework. I'm just trying to explain things better to my neice. I can't remember how to do complex

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Question 137998: I'm 27 and not in school anymore. I didn't want you to think that I'm cheating on homework. I'm just trying to explain things better to my neice. I can't remember how to do complex cubes. This is the one I'm stuck on.
I need to find all three of the complex cube roots of 27 in the A+Bi form.
Found 2 solutions by scott8148, ptaylor:
Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website!
x^3=27 __ x^3-27=0

factoring difference of cubes __ (x-3)(x^2+3x+9)=0

using quadratic formula __ x=(-3 +- sqrt[(3^2-(4)(1)(9)])/[(2)(1)] __ x=[-3 +- sqrt(-27)]/2
__ x=-3/2 +- [(3/2)sqrt(3)]i

one root is real (3) __ roots are 3+0i, (-3/2)+[(3/2)sqrt(3)]i, (-3/2)-[(3/2)sqrt(3)]i

Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website!
OK, Here's how you do it. Now we know, first of all, that all real numbers have one real root and two imaginary roots (Why? because inaginary roots occur in pairs)---27 is a real number
We also know, by inspection, that 3 is the real root of 27 because 3^3=27
If we let x^3=27, then x^3-27=0 This can be factored and we know what one factor is:
(x-3)(x^2 +?x+?)=0 In order to get the quadratic equation, we simply divide x-3 into x^3-27 and when we do that, we get x^2+3x+9. So now we have:
(x-3)(x^2+3x+9)=0
We'll solve x^2+3x+9 using the quadratic formula:

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Check my math
Hope this helps---ptaylor