Answer
$x \in\{-1- \sqrt 2i, -1+ \sqrt 2i, 2 \}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ 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, $a_{n}$.
------------------------
$f(x)=x^{3}-x-6$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 3, \pm 6$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm2, \pm3, \pm6$
b. Try for $x=2:$
$\begin{array}{lllll}
\underline{2}| & 1 & 0 & -1 & -6\\
& & 2 & 4 & 6\\
& -- & -- & -- & --\\
& 1 & 2 & 3 & |\underline{0}
\end{array}$
$2$ is a zero,
$f(x)=(x-2)(x^{2} +2x+3)$
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, $x^2+2x+3$, $x=\frac{-2 \pm \sqrt {2^2-4\times 1\times 3}}{2\times 1}=\frac{-2\pm 2\sqrt {2}i}{2}=-1\pm \sqrt 2i$
$x\displaystyle \in\{-1- \sqrt 2i, -1+ \sqrt 2i, 2 \}$
