Answer
The simplified form of the rational expression $\frac{{{x}^{3}}+2{{x}^{2}}+x}{{{x}^{2}}-4}\cdot \frac{{{x}^{2}}-x-2}{{{x}^{4}}+{{x}^{3}}}$ is $\frac{{{\left( x+1 \right)}^{2}}}{{{x}^{2}}\left( x+2 \right)}$.
Work Step by Step
$\frac{{{x}^{3}}+2{{x}^{2}}+x}{{{x}^{2}}-4}\cdot \frac{{{x}^{2}}-x-2}{{{x}^{4}}+{{x}^{3}}}$
$\frac{A}{B}\cdot \frac{C}{D}=\frac{AC}{BD}$
Multiplying the numerators and the denominators,
$\frac{{{x}^{3}}+2{{x}^{2}}+x}{{{x}^{2}}-4}\cdot \frac{{{x}^{2}}-x-2}{{{x}^{4}}+{{x}^{3}}}=\frac{\left( {{x}^{3}}+2{{x}^{2}}+x \right)\left( {{x}^{2}}-x-2 \right)}{\left( {{x}^{2}}-4 \right)\left( {{x}^{4}}+{{x}^{3}} \right)}$
$\begin{align}
& \left( {{x}^{3}}+2{{x}^{2}}+x \right)\left( {{x}^{2}}-x-2 \right)=x\left( {{x}^{2}}+2x+1 \right)\left( {{x}^{2}}-2x+x-2 \right) \\
& =x\left( {{x}^{2}}+x+x+1 \right)\left( x\left( x-2 \right)+\left( x-2 \right) \right) \\
& =x\left( x\left( x+1 \right)+\left( x+1 \right) \right)\left( x+1 \right)\left( x+2 \right) \\
& =x\left( x+1 \right)\left( x+1 \right)\left( x+1 \right)\left( x-2 \right)
\end{align}$
$\begin{align}
& \left( {{x}^{2}}-4 \right)\left( {{x}^{4}}+{{x}^{3}} \right)=\left( x+2 \right)\left( x-2 \right){{x}^{3}}\left( x+1 \right) \\
& ={{x}^{3}}\left( x+2 \right)\left( x-2 \right)\left( x+1 \right)
\end{align}$
So, the rational expression becomes,
$\frac{{{x}^{3}}+2{{x}^{2}}+x}{{{x}^{2}}-4}\cdot \frac{{{x}^{2}}-x-2}{{{x}^{4}}+{{x}^{3}}}=\frac{x\left( x+1 \right)\left( x+1 \right)\left( x+1 \right)\left( x-2 \right)}{{{x}^{3}}\left( x+2 \right)\left( x-2 \right)\left( x+1 \right)}$
Regroup and remove the factor equal to $1$,
$\begin{align}
& \frac{{{x}^{3}}+2{{x}^{2}}+x}{{{x}^{2}}-4}\cdot \frac{{{x}^{2}}-x-2}{{{x}^{4}}+{{x}^{3}}}=\frac{x{{\left( x+1 \right)}^{2}}\left( x+1 \right)\left( x-2 \right)}{\left( x+2 \right)\left( x-2 \right){{x}^{3}}\left( x+1 \right)} \\
& =1\cdot \frac{{{\left( x+1 \right)}^{2}}}{{{x}^{2}}\left( x+2 \right)} \\
& =\frac{{{\left( x+1 \right)}^{2}}}{{{x}^{2}}\left( x+2 \right)}
\end{align}$
