Simplify (4x^2-4x+1)/(8x^3-1)

$\frac{4 x^{2} - 4 x + 1}{8 x^{3} - 1}$

Factor using the perfect square rule.

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

$\frac{{(2 x)}^{2} - 4 x + 1}{8 x^{3} - 1}$

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

$\frac{{(2 x)}^{2} - 4 x + 1^{2}}{8 x^{3} - 1}$

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

$4 x = 2 \cdot (2 x) \cdot 1$

Rewrite the polynomial.

$\frac{{(2 x)}^{2} - 2 \cdot (2 x) \cdot 1 + 1^{2}}{8 x^{3} - 1}$

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

$\frac{{(2 x - 1)}^{2}}{8 x^{3} - 1}$

Simplify the denominator.

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

$\frac{{(2 x - 1)}^{2}}{{(2 x)}^{3} - 1}$

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

$\frac{{(2 x - 1)}^{2}}{{(2 x)}^{3} - 1^{3}}$

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 $b = 1$ .

$\frac{{(2 x - 1)}^{2}}{(2 x - 1) ({(2 x)}^{2} + 2 x \cdot 1 + 1^{2})}$

Simplify.

Apply the product rule to $2 x$ .

$\frac{{(2 x - 1)}^{2}}{(2 x - 1) (2^{2} x^{2} + 2 x \cdot 1 + 1^{2})}$

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

$\frac{{(2 x - 1)}^{2}}{(2 x - 1) (4 x^{2} + 2 x \cdot 1 + 1^{2})}$

Multiply $2$ by $1$ .

$\frac{{(2 x - 1)}^{2}}{(2 x - 1) (4 x^{2} + 2 x + 1^{2})}$

One to any power is one.

$\frac{{(2 x - 1)}^{2}}{(2 x - 1) (4 x^{2} + 2 x + 1)}$

Cancel the common factor of ${(2 x - 1)}^{2}$ and $2 x - 1$ .

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

$\frac{(2 x - 1) (2 x - 1)}{(2 x - 1) (4 x^{2} + 2 x + 1)}$

Cancel the common factors.

Cancel the common factor.

$\frac{(2 x - 1) (2 x - 1)}{(2 x - 1) (4 x^{2} + 2 x + 1)}$

Rewrite the expression.

$\frac{2 x - 1}{4 x^{2} + 2 x + 1}$

Simplify (4x^2-4x+1)/(8x^3-1)

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