Answer
$(-\infty,-1)\cup(-1,1)\cup(1,\infty)$
Work Step by Step
The denominator of the given function, $
f(x)=\dfrac{x^3-1}{x^2-1}
,$ cannot be $0$. Hence,
\begin{array}{l}\require{cancel}
x^2-1\ne0
\\\\
x^2\ne1
\\\\
x\ne\pm\sqrt{1}
\\\\
x\ne\pm1
.\end{array}
Hence, the domain is $
(-\infty,-1)\cup(-1,1)\cup(1,\infty)
.$