Simplify (8x^3-27)/(4x^2+6x+9)

$\frac{8 x^{3} - 27}{4 x^{2} + 6 x + 9}$

Step 1

Simplify the numerator.

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

$\frac{{(2 x)}^{3} - 27}{4 x^{2} + 6 x + 9}$

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

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

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

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

Simplify.

Apply the product rule to $2 x$ .

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

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

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

Multiply $3$ by $2$ .

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

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

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

Step 2

Cancel the common factor of $4 x^{2} + 6 x + 9$ .

Cancel the common factor.

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

Divide $2 x - 3$ by $1$ .

$2 x - 3$

Simplify (8x^3-27)/(4x^2+6x+9)

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