Factor 64x^6-1

$64 x^{6} - 1$

Step 1

Rewrite $64 x^{6}$ as ${(4 x^{2})}^{3}$ .

${(4 x^{2})}^{3} - 1$

Step 2

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

${(4 x^{2})}^{3} - 1^{3}$

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 $a = 4 x^{2}$ and $b = 1$ .

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

Step 4

Simplify.

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

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

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

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

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

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

Multiply $4$ by $1$ .

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

Step 5

Apply the product rule to $4 x^{2}$ .

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

Step 6

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

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

Step 7

Multiply the exponents in ${(x^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Multiply $2$ by $2$ .

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

Step 8

One to any power is one.

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

Step 9

Factor.

Rewrite $16 x^{4} + 4 x^{2} + 1$ in a factored form.

Rewrite the middle term.

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

Rearrange terms.

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

Factor first three terms by perfect square rule.

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

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4 x^{2} + 1$ and $b = 2 x$ .

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

Simplify.

Reorder terms.

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

Multiply $2$ by $- 1$ .

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

Reorder terms.

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

Remove unnecessary parentheses.

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

Factor 64x^6-1

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