Find all the real and complex roots of 8.

Your question is equivalent to solving for x  when x^3-8=0 .
Since we know that the real cube root of 8 is 2, we can divide by x-2  to get (x^3-8)/(x-2)=x^2+2x+4=0 (Use long division).
This equation can now be solved by completing the square:
x^2+2x=-4 
x^2+2x+1=-3
(x+1)^2=-3
x+1=+-isqrt(3)
X=-1+-isqrt(3)

Let the complex number z=8 . Rewrite this in polar form as z=8e^(i(0+2pi*n)) .
z^(1/3)=(8e^(i(0+2pi*n)))^(1/3)
z^(1/3)=2e^(i(0+(2pi*n)/3))
Here, theta becomes 3 distinct angles for all integers n . 0^@ , 120^@ , and 240^@ . The value at 0^@ is the real root, and the other two are complex roots. These can be represented as equal points around a circle of radius 2 in the complex plane.
The roots in cartesian coordinates are
8^(1/3)={2, -1+i sqrt(3), -1-i sqrt(3)}