Algebra Examples

$\frac{64 u^{3} + 8}{16 u^{2} - 4}$

Step 1

Cancel the common factor of $64 u^{3} + 8$ and $16 u^{2} - 4$ .

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Step 1.1

Factor $4$ out of $64 u^{3}$ .

$\frac{4 (16 u^{3}) + 8}{16 u^{2} - 4}$

Step 1.2

Factor $4$ out of $8$ .

$\frac{4 (16 u^{3}) + 4 (2)}{16 u^{2} - 4}$

Step 1.3

Factor $4$ out of $4 (16 u^{3}) + 4 (2)$ .

$\frac{4 (16 u^{3} + 2)}{16 u^{2} - 4}$

Step 1.4

Cancel the common factors.

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Step 1.4.1

Factor $4$ out of $16 u^{2}$ .

$\frac{4 (16 u^{3} + 2)}{4 (4 u^{2}) - 4}$

Step 1.4.2

Factor $4$ out of $- 4$ .

$\frac{4 (16 u^{3} + 2)}{4 (4 u^{2}) + 4 (- 1)}$

Step 1.4.3

Factor $4$ out of $4 (4 u^{2}) + 4 (- 1)$ .

$\frac{4 (16 u^{3} + 2)}{4 (4 u^{2} - 1)}$

Step 1.4.4

Cancel the common factor.

$\frac{4 (16 u^{3} + 2)}{4 (4 u^{2} - 1)}$

Step 1.4.5

Rewrite the expression.

$\frac{16 u^{3} + 2}{4 u^{2} - 1}$

Step 2

Simplify the numerator.

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Step 2.1

Factor $2$ out of $16 u^{3} + 2$ .

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Step 2.1.1

Factor $2$ out of $16 u^{3}$ .

$\frac{2 (8 u^{3}) + 2}{4 u^{2} - 1}$

Step 2.1.2

Factor $2$ out of $2$ .

$\frac{2 (8 u^{3}) + 2 (1)}{4 u^{2} - 1}$

Step 2.1.3

Factor $2$ out of $2 (8 u^{3}) + 2 (1)$ .

$\frac{2 (8 u^{3} + 1)}{4 u^{2} - 1}$

Step 2.2

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

$\frac{2 ({(2 u)}^{3} + 1)}{4 u^{2} - 1}$

Step 2.3

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

$\frac{2 ({(2 u)}^{3} + 1^{3})}{4 u^{2} - 1}$

Step 2.4

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 $a = 2 u$ and $b = 1$ .

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

Step 2.5

Simplify.

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Step 2.5.1

Apply the product rule to $2 u$ .

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

Step 2.5.2

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

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

Step 2.5.3

Multiply $2$ by $- 1$ .

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

Step 2.5.4

Multiply $- 2$ by $1$ .

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

Step 2.5.5

One to any power is one.

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

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

Step 3

Simplify the denominator.

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Step 3.1

Rewrite $4 u^{2}$ as ${(2 u)}^{2}$ .

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

Step 3.2

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

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

Step 3.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 u$ and $b = 1$ .

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

Step 4

Cancel the common factor of $2 u + 1$ .

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Step 4.1

Cancel the common factor.

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

Step 4.2

Rewrite the expression.

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