Trigonometry Examples

$f (x) = 8 x^{3} + \frac{1}{64}$

Step 1

Rewrite $8 x^{3}$ as ${(2 x)}^{3}$ .

$f (x) = {(2 x)}^{3} + \frac{1}{64}$

Step 2

Rewrite $1$ as $1^{3}$ .

$f (x) = {(2 x)}^{3} + \frac{1^{3}}{64}$

Step 3

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

$f (x) = {(2 x)}^{3} + \frac{1^{3}}{4^{3}}$

Step 4

Rewrite $\frac{1^{3}}{4^{3}}$ as ${(\frac{1}{4})}^{3}$ .

$f (x) = {(2 x)}^{3} + {(\frac{1}{4})}^{3}$

Step 5

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 $a = 2 x$ and $b = \frac{1}{4}$ .

$f (x) = (2 x + \frac{1}{4}) ({(2 x)}^{2} - 2 x \frac{1}{4} + {(\frac{1}{4})}^{2})$

Step 6

Simplify.

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Step 6.1

Apply the product rule to $2 x$ .

$f (x) = (2 x + \frac{1}{4}) (2^{2} x^{2} - 2 x \frac{1}{4} + {(\frac{1}{4})}^{2})$

Step 6.2

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

$f (x) = (2 x + \frac{1}{4}) (4 x^{2} - 2 x \frac{1}{4} + {(\frac{1}{4})}^{2})$

Step 6.3

Cancel the common factor of $2$ .

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Step 6.3.1

Factor $2$ out of $- (2 x)$ .

$f (x) = (2 x + \frac{1}{4}) (4 x^{2} + 2 (- (x)) (\frac{1}{4}) + {(\frac{1}{4})}^{2})$

Step 6.3.2

Factor $2$ out of $4$ .

$f (x) = (2 x + \frac{1}{4}) (4 x^{2} + 2 (- (x)) (\frac{1}{2 (2)}) + {(\frac{1}{4})}^{2})$

Step 6.3.3

Cancel the common factor.

$f (x) = (2 x + \frac{1}{4}) (4 x^{2} + 2 (- (x)) (\frac{1}{2 \cdot 2}) + {(\frac{1}{4})}^{2})$

Step 6.3.4

Rewrite the expression.

$f (x) = (2 x + \frac{1}{4}) (4 x^{2} - x \frac{1}{2} + {(\frac{1}{4})}^{2})$

Step 6.4

Combine $\frac{1}{2}$ and $x$ .

$f (x) = (2 x + \frac{1}{4}) (4 x^{2} - \frac{x}{2} + {(\frac{1}{4})}^{2})$

Step 6.5

Apply the product rule to $\frac{1}{4}$ .

$f (x) = (2 x + \frac{1}{4}) (4 x^{2} - \frac{x}{2} + \frac{1^{2}}{4^{2}})$

Step 6.6

One to any power is one.

$f (x) = (2 x + \frac{1}{4}) (4 x^{2} - \frac{x}{2} + \frac{1}{4^{2}})$

Step 6.7

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

$f (x) = (2 x + \frac{1}{4}) (4 x^{2} - \frac{x}{2} + \frac{1}{16})$