Algebra Examples

$x^{3} + \frac{1}{8}$

Step 1

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

$x^{3} + \frac{1^{3}}{8}$

Step 2

Rewrite $8$ as $2^{3}$ .

$x^{3} + \frac{1^{3}}{2^{3}}$

Step 3

Rewrite $\frac{1^{3}}{2^{3}}$ as ${(\frac{1}{2})}^{3}$ .

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

Step 4

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 $b = \frac{1}{2}$ .

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

Step 5

Simplify.

Tap for more steps...

Step 5.1

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

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

Step 5.2

Apply the product rule to $\frac{1}{2}$ .

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

Step 5.3

One to any power is one.

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

Step 5.4

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

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