Simplify (x^2-9)/(x^2+3x-18)*(x+6)/(x-5)

Rewrite ${\displaystyle 9}$ as ${\displaystyle 3^2}$.

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 ${\displaystyle b=3}$.

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 ${\displaystyle -18}$ and whose sum is ${\displaystyle 3}$.

Write the factored form using these integers.

Cancel the common factor of ${\displaystyle x+6}$.

Multiply ${\displaystyle \frac{(x+3)(x-3)}{x-3}}$ by ${\displaystyle \frac{1}{x-5}}$.

Cancel the common factor of ${\displaystyle x-3}$.