Answer
$2, -1 + i\sqrt 3, -1 - i\sqrt 3$
Work Step by Step
This question asks for ALL zeros of the function
Given $P(x) = x^3 - 8$
$P(x) = (x-2)(x^2 + 2x + 4)$
Set each polynomial equal to zero
$x-2 = 0$
$x = 2$
$x^2 + 2x + 4 = 0$
$x^2 + 2x + 1 = -3$
$x = -1 +/- i\sqrt 3$
Thus the solutions are $2, -1 + i\sqrt 3, -1 - i\sqrt 3$
