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

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

Step 1

Simplify the numerator.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\frac{{(2 x)}^{2} - 9}{8 x^{3} - 27}$

Rewrite $9$ as $3^{2}$ .

$\frac{{(2 x)}^{2} - 3^{2}}{8 x^{3} - 27}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 3$ .

$\frac{(2 x + 3) (2 x - 3)}{8 x^{3} - 27}$

Step 2

Simplify the denominator.

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

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

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

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

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

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

Simplify.

Apply the product rule to $2 x$ .

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

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

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

Multiply $3$ by $2$ .

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

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

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

Step 3

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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