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

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

Rewrite $27 x^{3}$ as ${(3 x)}^{3}$ .

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

Rewrite $8$ as $2^{3}$ .

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

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

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

Simplify.

Apply the product rule to $3 x$ .

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

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

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

Multiply $2$ by $3$ .

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

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

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

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

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

Rewrite $4$ as $2^{2}$ .

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

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

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

Reduce the expression $\frac{(3 x - 2) (9 x^{2} + 6 x + 4)}{(3 x + 2) (3 x - 2)}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

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

Related Questions

Support LCMGCF.com by disabling your adblocker.