Answer
$(-\frac{1}{2},1)U(3,\infty)$
Work Step by Step
Step 1. Factor the denominator $x^3-1=(x-1)(x^2+x+1)$. Identify boundary points (zeros and asymptotes) as $x=-\frac{1}{2},1,3$ which divide the x-axis into four intervals $(-\infty,-\frac{1}{2}),(-\frac{1}{2},1),(1,3),(3,\infty)$
Step 2. Choose test values in each interval $x=-1,0,2,4$ and test the inequality to get $False,\ True,\ False,\ True$
Step 3. Thus the solution intervals are $(-\frac{1}{2},1)U(3,\infty)$