Factor (x^6-1)

$(x^{6} - 1)$

Step 1

Remove parentheses.

$x^{6} - 1$

Step 2

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

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

Step 3

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

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

Step 4

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

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

Step 5

Simplify.

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

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

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

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

Multiply $x^{2}$ by $1$ .

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

Step 6

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

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

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

Multiply $2$ by $2$ .

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

Step 7

One to any power is one.

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

Step 8

Factor.

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

Rewrite the middle term.

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

Rearrange terms.

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

Factor first three terms by perfect square rule.

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

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

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

Simplify.

Reorder terms.

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

Reorder terms.

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

Remove unnecessary parentheses.

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

Factor (x^6-1)

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