Factor 8x^3-1

$8 x^{3} - 1$

Step 1

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

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

Step 2

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

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

Step 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 $b = 1$ .

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

Step 4

Simplify.

Apply the product rule to $2 x$ .

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

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

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

Multiply $2$ by $1$ .

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

One to any power is one.

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

Factor 8x^3-1

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