SOLUTION: Solve the equation in the complex number system. x^3-8=0 This one is stumping me. It doesn't appear that it can be factored out so I tried to use (x-1)=(x^2+X+8). I still c

Question 95502: Solve the equation in the complex number system.
x^3-8=0
This one is stumping me. It doesn't appear that it can be factored out so I tried to use (x-1)=(x^2+X+8). I still can not come up with an answer. Any help would be greatly appreciated.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve the equation in the complex number system.
x^3-8=0
It's the difference of cubes:
(x-2)(x^2+2x+4)=0
x=2 or x^2+2x+4=0
-------------------
Solve x^2+2x+4=0 using the quadratic formula:
x = [-2+-sqrt(4-4*1*4)/2
x = [-2+-sqrt(-12)]/2
x = [-2+-2sqrt(3)i]/2
x = -1+(sqrt(3))i or x=-1-(sqrt(3))i or x=2
====================
Cheers,
Stan H.

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