Simplify (x^2+2x+1)/((x^2-3x-18)/((x^2-1)/(x^2-7x+6)))

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 ${\displaystyle 6}$ and whose sum is ${\displaystyle -7}$.

Write the factored form using these integers.

Cancel the common factor.

Rewrite the expression.

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.

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

Raise ${\displaystyle x-6}$ to the power of ${\displaystyle 1}$.

Raise ${\displaystyle x-6}$ to the power of ${\displaystyle 1}$.

Use the power rule ${\displaystyle a^m{a}^n=a^{m+n}}$ to combine exponents.

Add ${\displaystyle 1}$ and ${\displaystyle 1}$.

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

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 ${\displaystyle b=1}$.

Multiply ${\displaystyle {(x+1)}^2}$ by ${\displaystyle x+1}$.

Add ${\displaystyle 2}$ and ${\displaystyle 1}$.