Basic Math Examples

$\frac{20 a}{4 a + 8 b} \div \frac{a^{3} - 8 b^{3}}{a^{2} - 4 b^{2}}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{20 a}{4 a + 8 b} \cdot \frac{a^{2} - 4 b^{2}}{a^{3} - 8 b^{3}}$

Step 2

Cancel the common factor of $20$ and $4 a + 8 b$ .

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Step 2.1

Factor $4$ out of $20 a$ .

$\frac{4 (5 a)}{4 a + 8 b} \cdot \frac{a^{2} - 4 b^{2}}{a^{3} - 8 b^{3}}$

Step 2.2

Cancel the common factors.

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Step 2.2.1

Factor $4$ out of $4 a$ .

$\frac{4 (5 a)}{4 (a) + 8 b} \cdot \frac{a^{2} - 4 b^{2}}{a^{3} - 8 b^{3}}$

Step 2.2.2

Factor $4$ out of $8 b$ .

$\frac{4 (5 a)}{4 (a) + 4 (2 b)} \cdot \frac{a^{2} - 4 b^{2}}{a^{3} - 8 b^{3}}$

Step 2.2.3

Factor $4$ out of $4 (a) + 4 (2 b)$ .

$\frac{4 (5 a)}{4 (a + 2 b)} \cdot \frac{a^{2} - 4 b^{2}}{a^{3} - 8 b^{3}}$

Step 2.2.4

Cancel the common factor.

$\frac{4 (5 a)}{4 (a + 2 b)} \cdot \frac{a^{2} - 4 b^{2}}{a^{3} - 8 b^{3}}$

Step 2.2.5

Rewrite the expression.

$\frac{5 a}{a + 2 b} \cdot \frac{a^{2} - 4 b^{2}}{a^{3} - 8 b^{3}}$

Step 3

Simplify the numerator.

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Step 3.1

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

$\frac{5 a}{a + 2 b} \cdot \frac{a^{2} - {(2 b)}^{2}}{a^{3} - 8 b^{3}}$

Step 3.2

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

$\frac{5 a}{a + 2 b} \cdot \frac{(a + 2 b) (a - (2 b))}{a^{3} - 8 b^{3}}$

Step 3.3

Multiply $2$ by $- 1$ .

$\frac{5 a}{a + 2 b} \cdot \frac{(a + 2 b) (a - 2 b)}{a^{3} - 8 b^{3}}$

Step 4

Simplify the denominator.

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Step 4.1

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

$\frac{5 a}{a + 2 b} \cdot \frac{(a + 2 b) (a - 2 b)}{a^{3} - {(2 b)}^{3}}$

Step 4.2

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 $a = a$ and $b = 2 b$ .

$\frac{5 a}{a + 2 b} \cdot \frac{(a + 2 b) (a - 2 b)}{(a - (2 b)) (a^{2} + a (2 b) + {(2 b)}^{2})}$

Step 4.3

Simplify.

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Step 4.3.1

Multiply $2$ by $- 1$ .

$\frac{5 a}{a + 2 b} \cdot \frac{(a + 2 b) (a - 2 b)}{(a - 2 b) (a^{2} + a (2 b) + {(2 b)}^{2})}$

Step 4.3.2

Rewrite using the commutative property of multiplication.

$\frac{5 a}{a + 2 b} \cdot \frac{(a + 2 b) (a - 2 b)}{(a - 2 b) (a^{2} + 2 a b + {(2 b)}^{2})}$

Step 4.3.3

Apply the product rule to $2 b$ .

$\frac{5 a}{a + 2 b} \cdot \frac{(a + 2 b) (a - 2 b)}{(a - 2 b) (a^{2} + 2 a b + 2^{2} b^{2})}$

Step 4.3.4

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

$\frac{5 a}{a + 2 b} \cdot \frac{(a + 2 b) (a - 2 b)}{(a - 2 b) (a^{2} + 2 a b + 4 b^{2})}$

Step 5

Simplify terms.

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Step 5.1

Cancel the common factor of $a + 2 b$ .

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Step 5.1.1

Cancel the common factor.

$\frac{5 a}{a + 2 b} \cdot \frac{(a + 2 b) (a - 2 b)}{(a - 2 b) (a^{2} + 2 a b + 4 b^{2})}$

Step 5.1.2

Rewrite the expression.

$5 a \frac{a - 2 b}{(a - 2 b) (a^{2} + 2 a b + 4 b^{2})}$

Step 5.2

Combine $5$ and $\frac{a - 2 b}{(a - 2 b) (a^{2} + 2 a b + 4 b^{2})}$ .

$a \frac{5 (a - 2 b)}{(a - 2 b) (a^{2} + 2 a b + 4 b^{2})}$

Step 5.3

Combine $a$ and $\frac{5 (a - 2 b)}{(a - 2 b) (a^{2} + 2 a b + 4 b^{2})}$ .

$\frac{a (5 (a - 2 b))}{(a - 2 b) (a^{2} + 2 a b + 4 b^{2})}$

Step 5.4

Cancel the common factor of $a - 2 b$ .

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Step 5.4.1

Cancel the common factor.

$\frac{a (5 (a - 2 b))}{(a - 2 b) (a^{2} + 2 a b + 4 b^{2})}$

Step 5.4.2

Rewrite the expression.

$\frac{a \cdot (5)}{a^{2} + 2 a b + 4 b^{2}}$

Step 5.5

Move $5$ to the left of $a$ .

$\frac{5 a}{a^{2} + 2 a b + 4 b^{2}}$