Answer
$x\in \{-4, -2, 1\}$
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, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2,\pm4,\pm8,$
$q:\qquad \pm 1, $
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm4,\pm8$
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,
$f(x)=(x-1)(x^2-2x-8)$
$x^2+2x-4x-8,$
$x(x+2)-4(x+2),$
$(x-4)(x+2),$
$f(x)=(x-1)(x+2)(x+4)$
$x\in \{-4, -2, 1\}$
