how about this
let y = x^3
then your equation becomes
y^2 - 7y - 8 = 0
(y-8)(y+1) = 0
so y = 8 or y = -1
then x^3 = 8 or x^3 = -1
x^3 - 8 = 0
(x-2)(x^2 + 2x + 4) = 0
this will give you the real root of x=2 and 2 imaginary roots from the quadratic
x^3 + 1 = 0
(x+1)(x^2 - x + 1) = 0
here x = -1 plus 2 more imaginary roots
Ohhhh duh!! I can't believe I totally overlooked that. Thanks Reiny!! :D
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