Pre-Algebra Examples

$\frac{x^{3} + 64}{x^{2} - 4 x + 16} \cdot \frac{11}{x^{2} - 16}$

Step 1

Simplify the numerator.

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Step 1.1

Rewrite $64$ as $4^{3}$ .

$\frac{x^{3} + 4^{3}}{x^{2} - 4 x + 16} \cdot \frac{11}{x^{2} - 16}$

Step 1.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 4$ .

$\frac{(x + 4) (x^{2} - x \cdot 4 + 4^{2})}{x^{2} - 4 x + 16} \cdot \frac{11}{x^{2} - 16}$

Step 1.3

Simplify.

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Step 1.3.1

Multiply $4$ by $- 1$ .

$\frac{(x + 4) (x^{2} - 4 x + 4^{2})}{x^{2} - 4 x + 16} \cdot \frac{11}{x^{2} - 16}$

Step 1.3.2

Raise $4$ to the power of $2$ .

$\frac{(x + 4) (x^{2} - 4 x + 16)}{x^{2} - 4 x + 16} \cdot \frac{11}{x^{2} - 16}$

Step 2

Simplify the denominator.

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Step 2.1

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

$\frac{(x + 4) (x^{2} - 4 x + 16)}{x^{2} - 4 x + 16} \cdot \frac{11}{x^{2} - 4^{2}}$

Step 2.2

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 $b = 4$ .

$\frac{(x + 4) (x^{2} - 4 x + 16)}{x^{2} - 4 x + 16} \cdot \frac{11}{(x + 4) (x - 4)}$

Step 3

Simplify terms.

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Step 3.1

Cancel the common factor of $x + 4$ .

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Step 3.1.1

Cancel the common factor.

$\frac{(x + 4) (x^{2} - 4 x + 16)}{x^{2} - 4 x + 16} \cdot \frac{11}{(x + 4) (x - 4)}$

Step 3.1.2

Rewrite the expression.

$\frac{x^{2} - 4 x + 16}{x^{2} - 4 x + 16} \cdot \frac{11}{x - 4}$

Step 3.2

Multiply $\frac{x^{2} - 4 x + 16}{x^{2} - 4 x + 16}$ by $\frac{11}{x - 4}$ .

$\frac{(x^{2} - 4 x + 16) \cdot 11}{(x^{2} - 4 x + 16) (x - 4)}$

Step 3.3

Cancel the common factor of $x^{2} - 4 x + 16$ .

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Step 3.3.1

Cancel the common factor.

$\frac{(x^{2} - 4 x + 16) \cdot 11}{(x^{2} - 4 x + 16) (x - 4)}$

Step 3.3.2

Rewrite the expression.

$\frac{11}{x - 4}$