Divide ((9-x^2)/(3x))÷((x^2+6x+9)/(3x))

$\frac{9 - x^{2}}{3 x} \div \frac{x^{2} + 6 x + 9}{3 x}$

To divide by a fraction, multiply by its reciprocal.

$\frac{9 - x^{2}}{3 x} \cdot \frac{3 x}{x^{2} + 6 x + 9}$

Cancel the common factor of $3 x$ .

Cancel the common factor.

$\frac{9 - x^{2}}{3 x} \cdot \frac{3 x}{x^{2} + 6 x + 9}$

Rewrite the expression.

$(9 - x^{2}) \frac{1}{x^{2} + 6 x + 9}$

Factor using the perfect square rule.

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

$(9 - x^{2}) \frac{1}{x^{2} + 6 x + 3^{2}}$

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

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

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

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

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

Multiply $9 - x^{2}$ by $\frac{1}{{(x + 3)}^{2}}$ .

$\frac{9 - x^{2}}{{(x + 3)}^{2}}$

Simplify the numerator.

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

$\frac{3^{2} - x^{2}}{{(x + 3)}^{2}}$

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

$\frac{(3 + x) (3 - x)}{{(x + 3)}^{2}}$

Cancel the common factor of $3 + x$ and ${(x + 3)}^{2}$ .

Reorder terms.

$\frac{(x + 3) (3 - x)}{{(x + 3)}^{2}}$

Cancel the common factors.

Factor $x + 3$ out of ${(x + 3)}^{2}$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{3 - x}{x + 3}$