How can I simplify x^3 - 27 completely?

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Homework Statement



I got down to here and didn't know where to go from here
http://www.wolframalpha.com/input/?...)(1-i^(-4/3))))(x+3/2(1+i^(2/3)(1-i^(-4/3))))
see the input value on that link if you don't feel like reading the equation below

(x-3) (x-3/2 (-1+i^(2/3) (1-i^(-4/3)))) (x+3/2 (1+i^(2/3) (1-i^(-4/3))))

now my question is how come wolfram tells me that the exact result is
(x-3) (x-3/2 ((-1)^(1/3) (1+(-1)^(1/3))-1)) (x+3/2 (1+(-1)^(1/3) (1+(-1)^(1/3))))
again click on the link to see the "exact result" if you don't want to read the equation like this

the two are not equal to each other by any means what's up with this?
Also the end result that I'm suppose to get to apparently is this

x^3+3/2 (1+e^((i pi)/3) (1+e^((i pi)/3))) x^2-3/2 (-1+e^((i pi)/3) (1+e^((i pi)/3))) x^2-3 x^2-9/4 (-1+e^((i pi)/3) (1+e^((i pi)/3))) (1+e^((i pi)/3) (1+e^((i pi)/3))) x-9/2 (1+e^((i pi)/3) (1+e^((i pi)/3))) x+9/2 (-1+e^((i pi)/3) (1+e^((i pi)/3))) x+27/4 (-1+e^((i pi)/3) (1+e^((i pi)/3))) (1+e^((i pi)/3) (1+e^((i pi)/3)))

the last alternate form on wolfram link above and currently I'm at the input form of x^3 - 27

(x-3) (x-3/2 (-1+i^(2/3) (1-i^(-4/3)))) (x+3/2 (1+i^(2/3) (1-i^(-4/3))))

any help in getting to that anser would be great as I don't know how to...
 
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The expressions are the same...wolfram replace i's with square roots of -1...but they are equivalent...
I don't really understand your question, you asked to simplify x^3-27...where do all these equations come from?
 
Did you mean "factorize"? x^3-27=x^3-3^3, so you can factor out x-3: x^3-3^3=(x-3)(x^2+3x+9). x^2+3x+9 can be factorized further as (x-u)(x-v) where u and v are the (complex) roots of x^2+3x+9=0.

ehild
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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