Answer
$x\displaystyle \in \left\{1, \frac{1- \sqrt 3i}{2}, \frac{1+\sqrt 3i}{2} \right\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and
$q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=x^{3}-2x^{2}+2x-1$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| & 1 & -2 & 2 & -1\\
& & 1 & -1 & 1\\
& -- & -- & -- & --\\
& 1 & -1 & 1 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(x^{2} -x+1)$
c. Solving for the trinomial using the quadratic formula for the quadratic equation of, $ax^2+bx+c$, $x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}$.
in this case, $x^2-x+1$, $x=\frac{1\pm \sqrt {(-1)^2-4 \times1\times1}}{2\times1}=\frac{1\pm \sqrt {3}i}{2}$
$x\displaystyle \in\left\{1, \frac{1- \sqrt 3i}{2}, \frac{1+\sqrt 3i}{2}\right\}$
