Factor 64c^3+1

$64 c^{3} + 1$

Step 1

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

${(4 c)}^{3} + 1$

Step 2

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

${(4 c)}^{3} + 1^{3}$

Step 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 $a = 4 c$ and $b = 1$ .

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

Step 4

Simplify.

Apply the product rule to $4 c$ .

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

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

$(4 c + 1) (16 c^{2} - (4 c) \cdot 1 + 1^{2})$

Multiply $4$ by $- 1$ .

$(4 c + 1) (16 c^{2} - 4 c \cdot 1 + 1^{2})$

Multiply $- 4$ by $1$ .

$(4 c + 1) (16 c^{2} - 4 c + 1^{2})$

One to any power is one.

$(4 c + 1) (16 c^{2} - 4 c + 1)$

Factor 64c^3+1

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