Factor (8x^3+1)

$(8 x^{3} + 1)$

Remove parentheses.

$8 x^{3} + 1$

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

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

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

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

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 2 x$ and $b = 1$ .

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

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 $- 1$ .

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

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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