Answer
$\frac{(x+3)(x-1)}{x+1}$.
Work Step by Step
First factor all numerators and denominators.
Factor $x^3+27$.
$\Rightarrow x^3+3^3$
Use the special formula $A^3+B^3=(A+B)(A^2-AB+B^2)$.
$\Rightarrow (x+3)(x^2-3x+3^2)$
Simplify.
$\Rightarrow (x+3)(x^2-3x+9)$
Factor $x^2-1$.
$\Rightarrow x^2-1^2$
Use the special formula $A^2-B^2=(A+B)(A-B)$.
$\Rightarrow (x+1)(x-1)$
Factor $x^2-2x+1$.
Use the special formula $(A-B)^2=A^2-2AB+B^2$.
$\Rightarrow (x-1)^2$
Back substitute all factors into the given fraction.
$\Rightarrow \frac{x^3+27}{x^2-1}\div\frac{x^2-3x+9}{x^2-2x+1}=\frac{(x+3)(x^2-3x+9)}{(x+1)(x-1)}\div\frac{x^2-3x+9}{(x-1)(x-1)}$
Invert the divisor and multiply.
$\Rightarrow \frac{(x+3)(x^2-3x+9)}{(x+1)(x-1)}\cdot\frac{(x-1)(x-1)}{x^2-3x+9}$
Cancel the common factors $x-1$, $x^2-3x+9$.
$\Rightarrow \frac{(x+3)(x-1)}{x+1}$.
