Simplify (x^2-9)/(x+1)*(x+1)/(x-3)

$\frac{x^{2} - 9}{x + 1} \cdot \frac{x + 1}{x - 3}$

Step 1

Simplify the numerator.

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

$\frac{x^{2} - 3^{2}}{x + 1} \cdot \frac{x + 1}{x - 3}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$\frac{(x + 3) (x - 3)}{x + 1} \cdot \frac{x + 1}{x - 3}$

Step 2

Reduce the expression by cancelling the common factors.

Cancel the common factor of $x - 3$ .

Factor $x - 3$ out of $(x + 3) (x - 3)$ .

$\frac{(x - 3) (x + 3)}{x + 1} \cdot \frac{x + 1}{x - 3}$

Cancel the common factor.

$\frac{(x - 3) (x + 3)}{x + 1} \cdot \frac{x + 1}{x - 3}$

Rewrite the expression.

$\frac{x + 3}{x + 1} \cdot (x + 1)$

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$\frac{x + 3}{x + 1} \cdot (x + 1)$

Rewrite the expression.

$x + 3$