Algebra Examples

$b^{3} + \frac{1}{27} a^{3}$

Step 1

Rewrite $\frac{1}{27} a^{3}$ as ${(\frac{1}{3} a)}^{3}$ .

$b^{3} + {(\frac{1}{3} a)}^{3}$

Step 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 $a = b$ and $b = \frac{1}{3} a$ .

$(b + \frac{1}{3} a) (b^{2} - b (\frac{1}{3} a) + {(\frac{1}{3} a)}^{2})$

Step 3

Simplify.

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Step 3.1

Combine $\frac{1}{3}$ and $a$ .

$(b + \frac{a}{3}) (b^{2} - b (\frac{1}{3} a) + {(\frac{1}{3} a)}^{2})$

Step 3.2

Rewrite using the commutative property of multiplication.

$(b + \frac{a}{3}) (b^{2} - 1 \cdot \frac{1}{3} b a + {(\frac{1}{3} a)}^{2})$

Step 3.3

Multiply $- 1$ by $\frac{1}{3}$ .

$(b + \frac{a}{3}) (b^{2} - 1 (\frac{1}{3}) b a + {(\frac{1}{3} a)}^{2})$

Step 3.4

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

$(b + \frac{a}{3}) (b^{2} - (\frac{1}{3}) b a + {(\frac{1}{3} a)}^{2})$

Step 3.5

Multiply $- 1$ by $\frac{1}{3}$ .

$(b + \frac{a}{3}) (b^{2} - \frac{1}{3} b a + {(\frac{1}{3} a)}^{2})$

Step 3.6

Combine $b$ and $\frac{1}{3}$ .

$(b + \frac{a}{3}) (b^{2} - \frac{b}{3} a + {(\frac{1}{3} a)}^{2})$

Step 3.7

Combine $a$ and $\frac{b}{3}$ .

$(b + \frac{a}{3}) (b^{2} - \frac{a b}{3} + {(\frac{1}{3} a)}^{2})$

Step 3.8

Combine $\frac{1}{3}$ and $a$ .

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

Step 3.9

Apply the product rule to $\frac{a}{3}$ .

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

Step 3.10

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

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

Step 3.11

Factor.

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Step 3.11.1

Factor $\frac{1}{9}$ out of each term.

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

Step 3.11.2

Remove unnecessary parentheses.

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