Simplify (27a^3-1)/(9a^2-6a+1)

$\frac{27 a^{3} - 1}{9 a^{2} - 6 a + 1}$

Step 1

Simplify the numerator.

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

$\frac{{(3 a)}^{3} - 1}{9 a^{2} - 6 a + 1}$

Rewrite $1$ as $1^{3}$ .

$\frac{{(3 a)}^{3} - 1^{3}}{9 a^{2} - 6 a + 1}$

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 $a = 3 a$ and $b = 1$ .

$\frac{(3 a - 1) ({(3 a)}^{2} + 3 a \cdot 1 + 1^{2})}{9 a^{2} - 6 a + 1}$

Simplify.

Apply the product rule to $3 a$ .

$\frac{(3 a - 1) (3^{2} a^{2} + 3 a \cdot 1 + 1^{2})}{9 a^{2} - 6 a + 1}$

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

$\frac{(3 a - 1) (9 a^{2} + 3 a \cdot 1 + 1^{2})}{9 a^{2} - 6 a + 1}$

Multiply $3$ by $1$ .

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1^{2})}{9 a^{2} - 6 a + 1}$

One to any power is one.

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1)}{9 a^{2} - 6 a + 1}$

Step 2

Factor using the perfect square rule.

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

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1)}{{(3 a)}^{2} - 6 a + 1}$

Rewrite $1$ as $1^{2}$ .

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1)}{{(3 a)}^{2} - 6 a + 1^{2}}$

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

$6 a = 2 \cdot (3 a) \cdot 1$

Rewrite the polynomial.

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1)}{{(3 a)}^{2} - 2 \cdot (3 a) \cdot 1 + 1^{2}}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 a$ and $b = 1$ .

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1)}{{(3 a - 1)}^{2}}$

Step 3

Cancel the common factors.

Factor $3 a - 1$ out of ${(3 a - 1)}^{2}$ .

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1)}{(3 a - 1) (3 a - 1)}$

Cancel the common factor.

$\frac{(3 a - 1) (9 a^{2} + 3 a + 1)}{(3 a - 1) (3 a - 1)}$

Rewrite the expression.

$\frac{9 a^{2} + 3 a + 1}{3 a - 1}$

Simplify (27a^3-1)/(9a^2-6a+1)

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