Simplify ((x^2)/9-1)/((x-6)/9+1/x)

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

Step 1

Multiply the numerator and denominator of the complex fraction by $9 x$ .

Multiply $\frac{\frac{x^{2}}{9} - 1}{\frac{x - 6}{9} + \frac{1}{x}}$ by $\frac{9 x}{9 x}$ .

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

Combine.

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

Step 2

Apply the distributive property.

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

Step 3

Simplify by cancelling.

Cancel the common factor of $9$ .

Factor $9$ out of $9 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $x$ by $x^{2}$ by adding the exponents.

Multiply $x$ by $x^{2}$ .

Raise $x$ to the power of $1$ .

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

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

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

Add $1$ and $2$ .

$\frac{x^{3} + 9 x \cdot - 1}{9 x \frac{x - 6}{9} + 9 x \frac{1}{x}}$

Cancel the common factor of $9$ .

Factor $9$ out of $9 x$ .

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

Cancel the common factor.

$\frac{x^{3} + 9 x \cdot - 1}{9 x \frac{x - 6}{9} + 9 x \frac{1}{x}}$

Rewrite the expression.

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

Cancel the common factor of $x$ .

Factor $x$ out of $9 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Simplify the numerator.

Factor $x$ out of $x^{3} + 9 x \cdot - 1$ .

Factor $x$ out of $x^{3}$ .

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

Factor $x$ out of $9 x \cdot - 1$ .

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

Factor $x$ out of $x \cdot x^{2} + x (9 \cdot - 1)$ .

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

Multiply $9$ by $- 1$ .

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

Rewrite $x^{2} - 9$ in a factored form.

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

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

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 $b = 3$ .

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

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

Step 5

Simplify the denominator.

Apply the distributive property.

$\frac{x (x + 3) (x - 3)}{x \cdot x + x \cdot - 6 + 9}$

Multiply $x$ by $x$ .

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

Move $- 6$ to the left of $x$ .

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

Factor using the perfect square rule.

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

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

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.

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

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

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

Step 6

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

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

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify ((x^2)/9-1)/((x-6)/9+1/x)

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