Recall that:
[tex]\begin{gathered} \text{If b}\ne0\text{ we get that:} \\ \frac{ab}{cb}=\frac{a}{c}\text{.} \end{gathered}[/tex]
Notice that:
[tex]\begin{gathered} 20n-16=4(5n-4), \\ 4n^2-4n^3=4n^2(1-n)=-4n^2(n-1), \\ 10n-10=10(n-1)\text{.} \end{gathered}[/tex]
Therefore:
[tex]\frac{20n-16}{5n-4}\cdot\frac{4n^2-4n^3}{10n-10}=\frac{4(5n-4)}{5n-4}\cdot\frac{-4n^2(n-1)}{10(n-1)}\text{.}[/tex]
Assuming that 5n-4≠0 and n-1≠0 we get:
[tex]\frac{4(5n-4)}{5n-4}\cdot\frac{-4n^2(n-1)}{10(n-1)}=4\cdot\frac{-4n^2}{10}\text{.}[/tex]
Simplifying the above result we get:
[tex]4\cdot\frac{-4n^2}{10}=-\frac{8n^2}{5}\text{.}[/tex]
Now, the restrictions are 5n-4≠0 and n-1≠0, therefore:
[tex]\begin{gathered} 5n-4\ne0\rightarrow5n\ne4\rightarrow n\ne\frac{4}{5}, \\ n-1\ne0\rightarrow n\ne1. \end{gathered}[/tex]
Answer:
[tex]\begin{gathered} -\frac{8n^2}{5}, \\ \text{For }n\ne1\text{ and }n\ne\frac{4}{5}. \end{gathered}[/tex]
