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