Answer
$2, \dfrac{1 + i\sqrt {3}}{2}, \dfrac{1- i\sqrt {3}}{2}$
Work Step by Step
Consider $P(x)=x^3-3x^2+3x-2=0$
In order to get the zeroes of the polynomial $P(x)$ we will have to solve as follows:
$x^3-3x^2+3x-2=(x-2)(x^2-x+1)=0$
This gives: $(x-2)(\dfrac{1\pm\sqrt{1-4}}{2})=0$
This can be re-written as:
$(x-2)(\dfrac{1\pm\sqrt{-3}}{2})=0$
$(x-2)(\dfrac{1\pm i\sqrt {3}}{2})=0 $
Thus the zeroes of $P(x)$ are: $2, \dfrac{1 + i\sqrt {3}}{2}, \dfrac{1- i\sqrt {3}}{2}$
