Algebra Examples

$\frac{108 a^{3} - 4}{36 a^{2} - 24 a + 4}$

Step 1

Cancel the common factor of $108 a^{3} - 4$ and $36 a^{2} - 24 a + 4$ .

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Step 1.1

Factor $4$ out of $108 a^{3}$ .

$\frac{4 (27 a^{3}) - 4}{36 a^{2} - 24 a + 4}$

Step 1.2

Factor $4$ out of $- 4$ .

$\frac{4 (27 a^{3}) + 4 \cdot - 1}{36 a^{2} - 24 a + 4}$

Step 1.3

Factor $4$ out of $4 (27 a^{3}) + 4 (- 1)$ .

$\frac{4 (27 a^{3} - 1)}{36 a^{2} - 24 a + 4}$

Step 1.4

Cancel the common factors.

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Step 1.4.1

Factor $4$ out of $36 a^{2}$ .

$\frac{4 (27 a^{3} - 1)}{4 (9 a^{2}) - 24 a + 4}$

Step 1.4.2

Factor $4$ out of $- 24 a$ .

$\frac{4 (27 a^{3} - 1)}{4 (9 a^{2}) + 4 (- 6 a) + 4}$

Step 1.4.3

Factor $4$ out of $4 (9 a^{2}) + 4 (- 6 a)$ .

$\frac{4 (27 a^{3} - 1)}{4 (9 a^{2} - 6 a) + 4}$

Step 1.4.4

Factor $4$ out of $4$ .

$\frac{4 (27 a^{3} - 1)}{4 (9 a^{2} - 6 a) + 4 (1)}$

Step 1.4.5

Factor $4$ out of $4 (9 a^{2} - 6 a) + 4 (1)$ .

$\frac{4 (27 a^{3} - 1)}{4 (9 a^{2} - 6 a + 1)}$

Step 1.4.6

Cancel the common factor.

$\frac{4 (27 a^{3} - 1)}{4 (9 a^{2} - 6 a + 1)}$

Step 1.4.7

Rewrite the expression.

$\frac{27 a^{3} - 1}{9 a^{2} - 6 a + 1}$

Step 2

Simplify the numerator.

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Step 2.1

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

$\frac{{(3 a)}^{3} - 1}{9 a^{2} - 6 a + 1}$

Step 2.2

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

$\frac{{(3 a)}^{3} - 1^{3}}{9 a^{2} - 6 a + 1}$

Step 2.3

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 $a = 3 a$ and $b = 1$ .

$\frac{(3 a - 1) ({(3 a)}^{2} + 3 a \cdot 1 + 1^{2})}{9 a^{2} - 6 a + 1}$

Step 2.4

Simplify.

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Step 2.4.1

Apply the product rule to $3 a$ .

$\frac{(3 a - 1) (3^{2} a^{2} + 3 a \cdot 1 + 1^{2})}{9 a^{2} - 6 a + 1}$

Step 2.4.2

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

$\frac{(3 a - 1) (9 a^{2} + 3 a \cdot 1 + 1^{2})}{9 a^{2} - 6 a + 1}$

Step 2.4.3

Multiply $3$ by $1$ .

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1^{2})}{9 a^{2} - 6 a + 1}$

Step 2.4.4

One to any power is one.

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1)}{9 a^{2} - 6 a + 1}$

Step 3

Factor using the perfect square rule.

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Step 3.1

Rewrite $9 a^{2}$ as ${(3 a)}^{2}$ .

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1)}{{(3 a)}^{2} - 6 a + 1}$

Step 3.2

Rewrite $1$ as $1^{2}$ .

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1)}{{(3 a)}^{2} - 6 a + 1^{2}}$

Step 3.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 a = 2 \cdot (3 a) \cdot 1$

Step 3.4

Rewrite the polynomial.

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

Step 3.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 a$ and $b = 1$ .

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

Step 4

Cancel the common factors.

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Step 4.1

Factor $3 a - 1$ out of ${(3 a - 1)}^{2}$ .

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

Step 4.2

Cancel the common factor.

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

Step 4.3

Rewrite the expression.

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