Simplify ((a-b)/(4a+4b))÷((a^2-b^2)/(a^2+6a+9))

$\frac{a - b}{4 a + 4 b} \div \frac{a^{2} - b^{2}}{a^{2} + 6 a + 9}$

To divide by a fraction, multiply by its reciprocal.

$\frac{a - b}{4 a + 4 b} \cdot \frac{a^{2} + 6 a + 9}{a^{2} - b^{2}}$

Factor $4$ out of $4 a + 4 b$ .

Factor $4$ out of $4 a$ .

$\frac{a - b}{4 (a) + 4 b} \cdot \frac{a^{2} + 6 a + 9}{a^{2} - b^{2}}$

Factor $4$ out of $4 b$ .

$\frac{a - b}{4 (a) + 4 (b)} \cdot \frac{a^{2} + 6 a + 9}{a^{2} - b^{2}}$

Factor $4$ out of $4 (a) + 4 (b)$ .

$\frac{a - b}{4 (a + b)} \cdot \frac{a^{2} + 6 a + 9}{a^{2} - b^{2}}$

Factor using the perfect square rule.

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

$\frac{a - b}{4 (a + b)} \cdot \frac{a^{2} + 6 a + 3^{2}}{a^{2} - b^{2}}$

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

$6 a = 2 \cdot a \cdot 3$

Rewrite the polynomial.

$\frac{a - b}{4 (a + b)} \cdot \frac{a^{2} + 2 \cdot a \cdot 3 + 3^{2}}{a^{2} - b^{2}}$

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

$\frac{a - b}{4 (a + b)} \cdot \frac{{(a + 3)}^{2}}{a^{2} - b^{2}}$

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

$\frac{a - b}{4 (a + b)} \cdot \frac{{(a + 3)}^{2}}{(a + b) (a - b)}$

Simplify terms.

Cancel the common factor of $a - b$ .

Factor $a - b$ out of $(a + b) (a - b)$ .

$\frac{a - b}{4 (a + b)} \cdot \frac{{(a + 3)}^{2}}{(a - b) (a + b)}$

Cancel the common factor.

$\frac{a - b}{4 (a + b)} \cdot \frac{{(a + 3)}^{2}}{(a - b) (a + b)}$

Rewrite the expression.

$\frac{1}{4 (a + b)} \cdot \frac{{(a + 3)}^{2}}{a + b}$

Multiply $\frac{1}{4 (a + b)}$ by $\frac{{(a + 3)}^{2}}{a + b}$ .

$\frac{{(a + 3)}^{2}}{4 (a + b) (a + b)}$

Raise $a + b$ to the power of $1$ .

$\frac{{(a + 3)}^{2}}{4 ({(a + b)}^{1} (a + b))}$

Raise $a + b$ to the power of $1$ .

$\frac{{(a + 3)}^{2}}{4 ({(a + b)}^{1} {(a + b)}^{1})}$

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

$\frac{{(a + 3)}^{2}}{4 {(a + b)}^{1 + 1}}$

Add $1$ and $1$ .

$\frac{{(a + 3)}^{2}}{4 {(a + b)}^{2}}$