Factorize the polynomial \(x^3 - 8\) over the complex numbers.

Step 1 ：Step 1: Recognize the expression as a difference of cubes. The general formula to factorize a difference of cubes is \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\).

Step 2 ：Step 2: In our case, \(a = x\) and \(b = 2\), so the factorized form is \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\).

Step 3 ：Step 3: The quadratic part \(x^2 + 2x + 4\) can be factored over the complex numbers. To find its roots, set it equal to zero and use the quadratic formula \(x = [-b \pm \sqrt{b^2 - 4ac}]/(2a)\).

Step 4 ：Step 4: Plug in the coefficients, \(a = 1\), \(b = 2\), and \(c = 4\), we get \(x = [-2 \pm \sqrt{(2)^2 - 4*1*4}]/(2*1)\), which simplifies to \(x = [-2 \pm \sqrt{-12}]/2\).

Step 5 ：Step 5: Simplify the expression using \(i\) for the square root of -1. We get \(x = -1 \pm \sqrt{3}i\).