Simplify ((x^2-9)/(x+1))÷((x^2-6x+9)/(7x+7))

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 ${\displaystyle b=3}$.

Factor ${\displaystyle 7}$ out of ${\displaystyle 7x}$.

Factor ${\displaystyle 7}$ out of ${\displaystyle 7}$.

Factor ${\displaystyle 7}$ out of ${\displaystyle 7\left(x\right)+7\left(1\right)}$.

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

Rewrite the polynomial.

Factor using the perfect square trinomial rule ${\displaystyle a^2-2ab+b^2={(a-b)}^2}$, where ${\displaystyle a=x}$ and ${\displaystyle b=3}$.

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

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

Multiply ${\displaystyle x+3}$ by ${\displaystyle \frac{7}{x-3}}$.

Move ${\displaystyle 7}$ to the left of ${\displaystyle x+3}$.