Precalculus Examples

$\frac{9 - x^{2}}{x^{3} - 27}$

Step 1

Simplify the numerator.

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Step 1.1

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

$\frac{3^{2} - x^{2}}{x^{3} - 27}$

Step 1.2

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 $b = x$ .

$\frac{(3 + x) (3 - x)}{x^{3} - 27}$

Step 2

Simplify the denominator.

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Step 2.1

Rewrite $27$ as $3^{3}$ .

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

Step 2.2

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

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

Step 2.3

Simplify.

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Step 2.3.1

Move $3$ to the left of $x$ .

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

Step 2.3.2

Raise $3$ to the power of $2$ .

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

Step 3

Reduce the expression by cancelling the common factors.

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Step 3.1

Cancel the common factor of $3 - x$ and $x - 3$ .

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Step 3.1.1

Rewrite $3$ as $- 1 (- 3)$ .

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

Step 3.1.2

Factor $- 1$ out of $- x$ .

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

Step 3.1.3

Factor $- 1$ out of $- 1 (- 3) - (x)$ .

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

Step 3.1.4

Reorder terms.

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

Step 3.1.5

Cancel the common factor.

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

Step 3.1.6

Rewrite the expression.

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

Step 3.2

Simplify the expression.

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Step 3.2.1

Move $- 1$ to the left of $3 + x$ .

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

Step 3.2.2

Move the negative in front of the fraction.

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