Answer
The inequality is valid for x-values between -0.5 and 1 and x-values more than 3 (not including them) i.e. $(-0.5,1)\cap(3,\infty)$.
Work Step by Step
Now, we find critical points by equating the numerator and denominator to zero:
$(3-x)^3(2x+1)=0$
$x^3-1=0$
There are three critical points:
$2x_1+1=0\rightarrow x_1=-\frac{1}{2}=-0.5$
$(x_2)^3-1=0\rightarrow (x_2)^3=1 \rightarrow \sqrt[3]{(x_2)^3}=\sqrt[3]1\rightarrow x_2=1$
$(3-x_3)^3=0\rightarrow\sqrt[3]{(3-x_3)^3}=\sqrt[3]0\rightarrow3-x_3=0\rightarrow x_3=3$
Next, we are going to take four values: one less than -0.5; one between -0.5 and 1; one between 1 and 3; and one more than 3 to test in the original equation and check if the inequality is true or not:
First test with a value less than -0.5:
$\dfrac{(3-(-1))^3(2(-1)+1)}{(-1)^3-1}<0$
$\dfrac{(4)^3(-2+1)}{-1-1}<0$
$\dfrac{64(-1)}{-2}<0$
$\dfrac{64}{2}<0$
$31<0 \rightarrow \text{ FALSE}$
Second test with a value between -0.5 and 1:
$\dfrac{(3-0)^3(2(0)+1)}{(0)^3-1}<0$
$\dfrac{3^3(1)}{-1}<0$
$-27<0 \rightarrow \text{ TRUE}$
Third test with a value between 1 and 3:
$\dfrac{(3-2)^3(2(2)+1)}{2^3-1}<0$
$\dfrac{(1)^3(5)}{8-1}<0$
$\dfrac{5}{7}<0 \rightarrow \text{ FALSE}$
Fourth test with a value more than 3:
$\dfrac{(3-4)^3(2(4)+1)}{4^3-1}<0$
$\dfrac{(-1)^3(9)}{64-1}<0$
$\dfrac{-9}{63}<0 \rightarrow \text{ TRUE}$
These tests show that the inequality $\dfrac{(3-x)^3(2x+1)}{x^3-1}<0$ is valid for values between -0.5 and 1 and values more than 3 (not including them) i.e. $(-0.5,1)\cap(3,\infty)$
