Factor s^6-1

$s^{6} - 1$

Step 1

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

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

Step 2

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

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

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

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

Step 4

Simplify.

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

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

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

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

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

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

Step 5

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

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

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

Multiply $2$ by $2$ .

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

Step 6

One to any power is one.

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

Step 7

Factor.

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

Rewrite the middle term.

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

Rearrange terms.

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

Factor first three terms by perfect square rule.

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

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

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

Simplify.

Reorder terms.

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

Reorder terms.

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

Remove unnecessary parentheses.

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

Factor s^6-1

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