x³ = 8
x³ - 8 = 0
a=x b=2
a³ - b³ = (a - b)(a² + ab + b²)
x³ - 8 = (x - 2)(x² + 2x + 2²)
(x - 2)(x² + 2x + 2²) = 0
(x - 2) = 0, (x² + 2x + 4) = 0
x = 2 [tex] \frac{-b +/- \sqrt{b x^{2} - 4ac} }{2a} = \frac{-2 +/- \sqrt{2 x^{2} - 4(1)(4)} }{2(1)} [/tex]
x = [tex] \frac{-2 +/- \sqrt{4 - 16} }{2} = \frac{-2 +/- \sqrt{-12} }{2} [/tex]
x = [tex] \frac{-2 +/- 2i \sqrt{3} }{2} = \frac{-1 +/- \sqrt{3}i }{1} [/tex]
Answer: the complex roots of 8 are: 2, -1 + √3i, -1 - √3i
