Basic Math Examples

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

Step 1

Simplify the numerator.

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Step 1.1

Rewrite $64$ as $4^{3}$ .

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

Step 1.2

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 $a = y$ and $b = 4$ .

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

Step 1.3

Simplify.

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Step 1.3.1

Multiply $4$ by $- 1$ .

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

Step 1.3.2

Raise $4$ to the power of $2$ .

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

Step 2

Simplify terms.

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Step 2.1

Factor $y$ out of $y^{3} - 4 y^{2} + 16 y$ .

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Step 2.1.1

Factor $y$ out of $y^{3}$ .

$\frac{(y + 4) (y^{2} - 4 y + 16)}{y \cdot y^{2} - 4 y^{2} + 16 y}$

Step 2.1.2

Factor $y$ out of $- 4 y^{2}$ .

$\frac{(y + 4) (y^{2} - 4 y + 16)}{y \cdot y^{2} + y (- 4 y) + 16 y}$

Step 2.1.3

Factor $y$ out of $16 y$ .

$\frac{(y + 4) (y^{2} - 4 y + 16)}{y \cdot y^{2} + y (- 4 y) + y \cdot 16}$

Step 2.1.4

Factor $y$ out of $y \cdot y^{2} + y (- 4 y)$ .

$\frac{(y + 4) (y^{2} - 4 y + 16)}{y (y^{2} - 4 y) + y \cdot 16}$

Step 2.1.5

Factor $y$ out of $y (y^{2} - 4 y) + y \cdot 16$ .

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

Step 2.2

Cancel the common factor of $y^{2} - 4 y + 16$ .

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Step 2.2.1

Cancel the common factor.

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

Step 2.2.2

Rewrite the expression.

$\frac{y + 4}{y}$