Basic Math Examples

$\frac{z^{3} - 8}{z^{3} + 8} \div \frac{z^{2} - 4}{z^{2} - 2 z + 4}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{z^{3} - 8}{z^{3} + 8} \cdot \frac{z^{2} - 2 z + 4}{z^{2} - 4}$

Step 2

Simplify the numerator.

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Step 2.1

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

$\frac{z^{3} - 2^{3}}{z^{3} + 8} \cdot \frac{z^{2} - 2 z + 4}{z^{2} - 4}$

Step 2.2

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

$\frac{(z - 2) (z^{2} + z \cdot 2 + 2^{2})}{z^{3} + 8} \cdot \frac{z^{2} - 2 z + 4}{z^{2} - 4}$

Step 2.3

Simplify.

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Step 2.3.1

Move $2$ to the left of $z$ .

$\frac{(z - 2) (z^{2} + 2 \cdot z + 2^{2})}{z^{3} + 8} \cdot \frac{z^{2} - 2 z + 4}{z^{2} - 4}$

Step 2.3.2

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

$\frac{(z - 2) (z^{2} + 2 z + 4)}{z^{3} + 8} \cdot \frac{z^{2} - 2 z + 4}{z^{2} - 4}$

Step 3

Simplify the denominator.

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Step 3.1

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

$\frac{(z - 2) (z^{2} + 2 z + 4)}{z^{3} + 2^{3}} \cdot \frac{z^{2} - 2 z + 4}{z^{2} - 4}$

Step 3.2

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

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

Step 3.3

Simplify.

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Step 3.3.1

Multiply $2$ by $- 1$ .

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

Step 3.3.2

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

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

Step 4

Simplify terms.

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Step 4.1

Cancel the common factor of $z^{2} - 2 z + 4$ .

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Step 4.1.1

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

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

Step 4.1.2

Cancel the common factor.

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

Step 4.1.3

Rewrite the expression.

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

Step 4.2

Multiply $\frac{(z - 2) (z^{2} + 2 z + 4)}{z + 2}$ by $\frac{1}{z^{2} - 4}$ .

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

Step 5

Simplify the denominator.

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Step 5.1

Rewrite $4$ as $2^{2}$ .

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

Step 5.2

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

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

Step 5.3

Combine exponents.

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Step 5.3.1

Raise $z + 2$ to the power of $1$ .

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

Step 5.3.2

Raise $z + 2$ to the power of $1$ .

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

Step 5.3.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.3.4

Add $1$ and $1$ .

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

Step 6

Cancel the common factor of $z - 2$ .

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Step 6.1

Cancel the common factor.

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

Step 6.2

Rewrite the expression.

$\frac{z^{2} + 2 z + 4}{{(z + 2)}^{2}}$