Answer
$(-\frac{1}{2},1)\cup(3,\infty)$.
Work Step by Step
Step 1. For $\frac{(3-x)^3(2x+1)}{(x-1)(x^2+x+1)}\lt0$, identify boundary points as $x=-\frac{1}{2},1,3$.
Step 2. Form intervals $(-\infty,-\frac{1}{2}),(-\frac{1}{2},1),(1,3),(3,\infty)$.
Step 3. Choose test values for each interval $x=-1,0,2,4$.
Step 4. Test the inequality to get results: $False,\ True,\ False,\ True$
Step 5. We have the solution $(-\frac{1}{2},1)\cup(3,\infty)$.
