Simplify (9-12x+4x^2)/(27-8x^3)

$\frac{9 - 12 x + 4 x^{2}}{27 - 8 x^{3}}$

Factor using the perfect square rule.

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

$\frac{3^{2} - 12 x + 4 x^{2}}{27 - 8 x^{3}}$

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

$\frac{3^{2} - 12 x + {(2 x)}^{2}}{27 - 8 x^{3}}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 x = 2 \cdot 3 \cdot (2 x)$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3$ and $b = 2 x$ .

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

Simplify the denominator.

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

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

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

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

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

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

Simplify.

Multiply $2$ by $- 1$ .

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

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

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

Multiply $2$ by $3$ .

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

Apply the product rule to $2 x$ .

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

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

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

Cancel the common factor of ${(3 - 2 x)}^{2}$ and $3 - 2 x$ .

Factor $3 - 2 x$ out of ${(3 - 2 x)}^{2}$ .

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Simplify (9-12x+4x^2)/(27-8x^3)

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