Answer
$x\displaystyle \in\left\{\frac{1- \sqrt {3}i}{2}. \frac{1+ \sqrt {3}i}{2}, 2 \right\}$
Work Step by Step
See The Rational Zero Theorem:
If $\displaystyle \frac{p}{q}$ is a zero of the polynomial $f$ with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
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$f(x)=x^{3}-3x^{2}+3x-2$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm2$
b. Try for $x=2:$
$\begin{array}{lllll}
\underline{2}| & 1 & -3 & 3 & -2\\
& & 2 & -2 & 2\\
& -- & -- & -- & --\\
& 1 & -1 & 1 & |\underline{0}
\end{array}$
$2$ is a zero,
$f(x)=(x-2)(x^{2} -x+1)$
c. Solve 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, $x^2-x+1$, $x=\frac{1 \pm \sqrt {-1^2-4\times 1\times 1}}{2\times 1}=\frac{1\pm \sqrt {3}i}{2}$
$x\displaystyle \in\left\{\frac{1- \sqrt {3}i}{2}. \frac{1+ \sqrt {3}i}{2}, 2 \right\}$
