Factor 64x^3+1=0

$64 x^{3} + 1 = 0$

Rewrite $64 x^{3}$ as ${(4 x)}^{3}$ .

${(4 x)}^{3} + 1 = 0$

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

${(4 x)}^{3} + 1^{3} = 0$

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 $a = 4 x$ and $b = 1$ .

$(4 x + 1) ({(4 x)}^{2} - (4 x) \cdot 1 + 1^{2}) = 0$

Simplify.

Apply the product rule to $4 x$ .

$(4 x + 1) (4^{2} x^{2} - (4 x) \cdot 1 + 1^{2}) = 0$

Raise $4$ to the power of $2$ .

$(4 x + 1) (16 x^{2} - (4 x) \cdot 1 + 1^{2}) = 0$

Multiply $4$ by $- 1$ .

$(4 x + 1) (16 x^{2} - 4 x \cdot 1 + 1^{2}) = 0$

Multiply $- 4$ by $1$ .

$(4 x + 1) (16 x^{2} - 4 x + 1^{2}) = 0$

One to any power is one.

$(4 x + 1) (16 x^{2} - 4 x + 1) = 0$

Factor 64x^3+1=0

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