Simplify (8y^3-216)/(2y-6)

$\frac{8 y^{3} - 216}{2 y - 6}$

Cancel the common factor of $8 y^{3} - 216$ and $2 y - 6$ .

Factor $2$ out of $8 y^{3}$ .

$\frac{2 (4 y^{3}) - 216}{2 y - 6}$

Factor $2$ out of $- 216$ .

$\frac{2 (4 y^{3}) + 2 \cdot - 108}{2 y - 6}$

Factor $2$ out of $2 (4 y^{3}) + 2 (- 108)$ .

$\frac{2 (4 y^{3} - 108)}{2 y - 6}$

Cancel the common factors.

Factor $2$ out of $2 y$ .

$\frac{2 (4 y^{3} - 108)}{2 (y) - 6}$

Factor $2$ out of $- 6$ .

$\frac{2 (4 y^{3} - 108)}{2 y + 2 \cdot - 3}$

Factor $2$ out of $2 y + 2 \cdot - 3$ .

$\frac{2 (4 y^{3} - 108)}{2 (y - 3)}$

Cancel the common factor.

$\frac{2 (4 y^{3} - 108)}{2 (y - 3)}$

Rewrite the expression.

$\frac{4 y^{3} - 108}{y - 3}$

Simplify the numerator.

Factor $4$ out of $4 y^{3} - 108$ .

Factor $4$ out of $4 y^{3}$ .

$\frac{4 (y^{3}) - 108}{y - 3}$

Factor $4$ out of $- 108$ .

$\frac{4 y^{3} + 4 \cdot - 27}{y - 3}$

Factor $4$ out of $4 y^{3} + 4 \cdot - 27$ .

$\frac{4 (y^{3} - 27)}{y - 3}$

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

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

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

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

Simplify.

Move $3$ to the left of $y$ .

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

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

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

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

Simplify terms.

Cancel the common factor of $y - 3$ .

Cancel the common factor.

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

Divide $4 (y^{2} + 3 y + 9)$ by $1$ .

$4 (y^{2} + 3 y + 9)$

Apply the distributive property.

$4 y^{2} + 4 (3 y) + 4 \cdot 9$

Simplify.

Multiply $3$ by $4$ .

$4 y^{2} + 12 y + 4 \cdot 9$

Multiply $4$ by $9$ .

$4 y^{2} + 12 y + 36$

Simplify (8y^3-216)/(2y-6)

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