Simplify ((a-b)/(7a+7b))÷((a^2-b^2)/(a^2+8a+16))

$\frac{a - b}{7 a + 7 b} \div \frac{a^{2} - b^{2}}{a^{2} + 8 a + 16}$

To divide by a fraction, multiply by its reciprocal.

$\frac{a - b}{7 a + 7 b} \cdot \frac{a^{2} + 8 a + 16}{a^{2} - b^{2}}$

Factor $7$ out of $7 a + 7 b$ .

Factor $7$ out of $7 a$ .

$\frac{a - b}{7 (a) + 7 b} \cdot \frac{a^{2} + 8 a + 16}{a^{2} - b^{2}}$

Factor $7$ out of $7 b$ .

$\frac{a - b}{7 (a) + 7 (b)} \cdot \frac{a^{2} + 8 a + 16}{a^{2} - b^{2}}$

Factor $7$ out of $7 (a) + 7 (b)$ .

$\frac{a - b}{7 (a + b)} \cdot \frac{a^{2} + 8 a + 16}{a^{2} - b^{2}}$

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$\frac{a - b}{7 (a + b)} \cdot \frac{a^{2} + 8 a + 4^{2}}{a^{2} - b^{2}}$

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

$8 a = 2 \cdot a \cdot 4$

Rewrite the polynomial.

$\frac{a - b}{7 (a + b)} \cdot \frac{a^{2} + 2 \cdot a \cdot 4 + 4^{2}}{a^{2} - b^{2}}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = a$ and $b = 4$ .

$\frac{a - b}{7 (a + b)} \cdot \frac{{(a + 4)}^{2}}{a^{2} - b^{2}}$

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 $b = b$ .

$\frac{a - b}{7 (a + b)} \cdot \frac{{(a + 4)}^{2}}{(a + b) (a - b)}$

Simplify terms.

Cancel the common factor of $a - b$ .

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

$\frac{a - b}{7 (a + b)} \cdot \frac{{(a + 4)}^{2}}{(a - b) (a + b)}$

Cancel the common factor.

$\frac{a - b}{7 (a + b)} \cdot \frac{{(a + 4)}^{2}}{(a - b) (a + b)}$

Rewrite the expression.

$\frac{1}{7 (a + b)} \cdot \frac{{(a + 4)}^{2}}{a + b}$

Multiply $\frac{1}{7 (a + b)}$ by $\frac{{(a + 4)}^{2}}{a + b}$ .

$\frac{{(a + 4)}^{2}}{7 (a + b) (a + b)}$

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

$\frac{{(a + 4)}^{2}}{7 ({(a + b)}^{1} (a + b))}$

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

$\frac{{(a + 4)}^{2}}{7 ({(a + b)}^{1} {(a + b)}^{1})}$

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

$\frac{{(a + 4)}^{2}}{7 {(a + b)}^{1 + 1}}$

Add $1$ and $1$ .

$\frac{{(a + 4)}^{2}}{7 {(a + b)}^{2}}$