Precalculus Examples

$\frac{8 y^{3} - 27}{2 y - 3}$

Step 1

Simplify the numerator.

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Step 1.1

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

$\frac{{(2 y)}^{3} - 27}{2 y - 3}$

Step 1.2

Rewrite $27$ as $3^{3}$ .

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

Step 1.3

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 $a = 2 y$ and $b = 3$ .

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

Step 1.4

Simplify.

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Step 1.4.1

Apply the product rule to $2 y$ .

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

Step 1.4.2

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

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

Step 1.4.3

Multiply $3$ by $2$ .

$\frac{(2 y - 3) (4 y^{2} + 6 y + 3^{2})}{2 y - 3}$

Step 1.4.4

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

$\frac{(2 y - 3) (4 y^{2} + 6 y + 9)}{2 y - 3}$

Step 2

Cancel the common factor of $2 y - 3$ .

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Step 2.1

Cancel the common factor.

$\frac{(2 y - 3) (4 y^{2} + 6 y + 9)}{2 y - 3}$

Step 2.2

Divide $4 y^{2} + 6 y + 9$ by $1$ .

$4 y^{2} + 6 y + 9$