Question 95502
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
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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
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Cheers,
Stan H.