Answer
$x\in \{-2, 1, 4\}$
Work Step by Step
See The Rational Zero Theorem:
... If $\displaystyle \frac{p}{q}$ 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, $a_{n}$.
------------------------
$f(x)=x^3-3x^2-6x+8$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 4, \pm 8$
$q:\qquad \pm 1, $
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm4, \pm 8$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| & 1 & -3 & -6 & 8\\
& & 1 & -2 & -8\\
& -- & -- & -- & --\\
& 1 & -2 & -8 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(x^2-2x+8)$
c. Factorize the trinomial factor $(x^2-2x+8)$
(find two factors of $1(8)=8$ whose sum is $-2):$
$(2$ and $-4$)
$x^{2} -2x+8=x^{2} +2x-4x-8 \quad$...factor in pairs ...
$x(x+2)-4(x+2)=(x+2)(x-4)$
$f(x)=(x-1)(x-4)(x+2)$
The zeros of f satisfy $f(x)=0$
$x\in \{-2, 1, 4\}$
