Answer
The inequality is valid for x-values between -1 and $\frac{2}{3}$ and x-values more than 2 (not including them) i.e. $(-1,\frac{2}{3})\cap(2,\infty)$
Work Step by Step
Now, we find critical points by equating the numerator and denominator to zero:
$(2-x)^3(3x-2)=0$
$x^3+1=0$
There are three critical points:
$(x_1)^3+1=0\rightarrow (x_1)^3=-1 \rightarrow \sqrt[3]{(x_1)^3}=\sqrt[3]-1\rightarrow x_1=-1$
$3x_2-2=0\rightarrow x_2=\frac{2}{3}$
$(2-x_3)^3=0\rightarrow\sqrt[3]{(2-x_3)^3}=\sqrt[3]0\rightarrow2-x_3=0\rightarrow x_3=2$
Next, we are going to take four values: one less than -1; one between -1 and $\frac{2}{3}$; one between $\frac{2}{3}$ and 2; and one more than 2 to test in the original equation and check if the inequality is true or not:
First test with a value less than -1:
$\dfrac{(2-(-2))^3(3(-2)-2)}{(-2)^3+1}<0$
$\dfrac{(4)^3(-6-2)}{-8+1}<0$
$\dfrac{64(-8)}{-7}<0$
$\dfrac{512}{7}<0 \rightarrow \text{ FALSE}$
Second test with a value between -1 and $\frac{2}{3}$:
$\dfrac{(2-0)^3(3(0)-2)}{(0)^3+1}<0$
$\dfrac{(2)^3(-2)}{1}<0$
$8(-2)<0$
$-16<0 \rightarrow \text{ TRUE}$
Third test with a value between $\frac{2}{3}$ and 2:
$\dfrac{(2-1)^3(3(1)-2)}{1^3+1}<0$
$\dfrac{1^3(3-2)}{1+1}<0$
$\dfrac{1(1)}{2}<0$
$0.5<0 \rightarrow \text{ FALSE}$
Fourth test with a value more than 2:
$\dfrac{(2-3)^3(3(3)-2)}{3^3+1}<0$
$\dfrac{(-1)^3(9-2)}{27+1}<0$
$\dfrac{-1(7)}{28}<0$
$-\dfrac{1}{4}<0 \rightarrow \text{ TRUE}$
These tests show that the inequality $\dfrac{(2-x)^3(3x-2)}{x^3+1}<0$ is valid for values between -1 and $\frac{2}{3}$ and values more than 2 (not including them) i.e. $(-1,\frac{2}{3})\cap(2,\infty)$
