Answer
$\underline{\left( -1,0 \right)\cup \left( 3,\infty \right)}$.
Work Step by Step
Rewrite the given inequality by subtracting $2{{x}^{2}}$ from both the sides of the inequality
$\begin{align}
& {{x}^{3}}-3x-2{{x}^{2}}>2{{x}^{2}}-2{{x}^{2}} \\
& {{x}^{3}}-3x-2{{x}^{2}}>0 \\
\end{align}$
Consider the “related” function,
$\begin{align}
& f\left( x \right)=0 \\
& {{x}^{3}}-3x-2{{x}^{2}}=0
\end{align}$
Since $f\left( x \right)$ is a third degree equation, $f\left( x \right)$ has three zeros.
Factor out $x$ from the above equation.
$\begin{align}
& x\left( {{x}^{2}}-3-2x \right)=0 \\
& x\left( {{x}^{2}}-2x-3 \right)=0 \\
& x\left( {{x}^{2}}-3x+x-3 \right)=0 \\
& x\left( x\left( x-3 \right)+\left( x-3 \right) \right)=0
\end{align}$
Apply the zero product rule,
$\begin{align}
& x\left( x+1 \right)\left( x-3 \right)=0 \\
& x=0\text{ or }x+1=0\text{ or }x-3=0 \\
& x=0\text{ or }x=-1\text{ or }x=3 \\
\end{align}$
Thus the zeros of $f\left( x \right)$are $0,-1\text{ and }3$.
Select one test value from each interval and determine the sign of $f\left( x \right)$ over that interval.
For interval A:
$\begin{align}
& f\left( -2 \right)=-2\left( -2+1 \right)\left( -2-3 \right) \\
& =\left( -2 \right)\left( -1 \right)\left( -5 \right) \\
& =-10
\end{align}$
Thus $f\left( -2 \right)$ is negative.
For interval B:
$\begin{align}
& f\left( -\frac{1}{2} \right)=-\frac{1}{2}\left( -\frac{1}{2}+1 \right)\left( -\frac{1}{2}-3 \right) \\
& =\left( -\frac{1}{2} \right)\cdot \left( \frac{1}{2} \right)\cdot \left( -\frac{7}{2} \right) \\
& =\frac{7}{8}
\end{align}$
Thus $f\left( -\frac{1}{2} \right)$ is positive.
For interval C:
$\begin{align}
& f\left( 1 \right)=1\left( 1+1 \right)\left( 1-3 \right) \\
& =1\cdot 2\cdot \left( -2 \right) \\
& =-4
\end{align}$
Thus $f\left( 1 \right)$ is negative.
For interval D:
$\begin{align}
& f\left( \frac{7}{2} \right)=\frac{7}{2}\left( \frac{7}{2}+1 \right)\left( \frac{7}{2}-3 \right) \\
& =\frac{7}{2}\cdot \frac{9}{2}\cdot \frac{1}{2} \\
& =\frac{63}{8}
\end{align}$
Thus $f\left( \frac{7}{2} \right)$ is positive.
Indicate the sign of $f\left( x \right)$ on the number line.
Look for the solution of ${{x}^{3}}-3x-2{{x}^{2}}>0$
The calculations above indicate that $f\left( x \right)$ is positive for any number in intervals B and D.
Therefore the solution set of the original inequality is
$\left( -1,0 \right)\cup \left( 3,\infty \right)$.
