Simplify (x^2-1)/(x^2-6x+5)*(3x-15)/(6x)

Rewrite ${\displaystyle 1}$ as ${\displaystyle 1^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=x}$ and ${\displaystyle b=1}$.

Consider the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$. In this case, whose product is ${\displaystyle 5}$ and whose sum is ${\displaystyle -6}$.

Write the factored form using these integers.

Factor ${\displaystyle 3}$ out of ${\displaystyle 3x-15}$.

Cancel the common factor of ${\displaystyle x-5}$.

Multiply ${\displaystyle \frac{(x+1)(x-1)}{x-1}}$ by ${\displaystyle \frac{3}{6x}}$.

Cancel the common factor of ${\displaystyle x-1}$.

Cancel the common factor of ${\displaystyle 3}$ and ${\displaystyle 6}$.