We have the following rational expression:
[tex]\frac{(a-1)^2(a-5)}{(a-5)^2(a-1)}[/tex]And we have to reduce the expression to the lowest terms.
1. To achieve that, we can see that:
• There are two common factors:
[tex](a-1)\text{ and }(a-5)[/tex]2. Now, we know that:
[tex]\begin{gathered} a^2=a*a \\ \\ \text{ And also, we have that:} \\ \\ \frac{a}{a}=\frac{b}{b}=1 \end{gathered}[/tex]3. And we can apply the same rule for the given case as follows:
[tex]\begin{gathered} \frac{(a-1)^{2}(a-5)}{(a-5)^{2}(a-1)}\Rightarrow(a-1)^2=(a-1)(a-1),(a-5)^2=(a-5)(a-5) \\ \\ \text{ Then we have:} \\ \\ \frac{(a-1)(a-1)(a-5)}{(a-5)(a-5)(a-1)} \end{gathered}[/tex]4. And we can rewrite the expression as follows:
[tex]\begin{gathered} \frac{(a-1)(a-1)(a-5)}{(a-5)(a-5)(a-1)}=\frac{(a-1)}{(a-1)}\frac{(a-5)}{(a-5)}\frac{(a-1)}{(a-5)}\Rightarrow\frac{(a-1)}{(a-1)}=1,\frac{(a-5)}{(a-5)}=1 \\ \\ \text{ Then we have:} \\ \\ \frac{(a-1)(a-1)(a-5)}{(a-5)(a-5)(a-1)}=(1)(1)\frac{(a-1)}{(a-5)}=\frac{(a-1)}{(a-5)} \\ \\ \end{gathered}[/tex]Therefore, the rational expression of the lowest terms is:
[tex]\frac{a-1}{a-5}[/tex]5. We can determine that the variable restrictions for the original expression can be found as follows:
[tex]\begin{gathered} \frac{(a-1)^{2}(a-5)}{(a-5)^{2}(a-1)} \\ \\ \text{ If a = 5, then we would have:} \\ \\ \frac{(a-1)^2(5-5)}{(5-5)^2(a-1)}=\frac{0}{0}\rightarrow\text{ This is NOT defined. Then a}\ne5 \end{gathered}[/tex][tex]\begin{gathered} \frac{(a-1)^{2}(a-5)}{(a-5)^{2}(a-1)} \\ \\ \text{ If a = 1, then we have:} \\ \\ \frac{(1-1)^2(a-5)}{(a-5)^2(1-1)}=\frac{0}{0}\rightarrow\text{ This is NOT defined. Then a}\ne1 \end{gathered}[/tex]Therefore, in summary, we have that:
• The rational expression in the lowest terms is:
[tex]\frac{a-1}{a-5}[/tex]• The variable restrictions for the original expression:
[tex]a\ne1,5[/tex]
