Factor 1-p^9

$1 - p^{9}$

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

$1^{3} - p^{9}$

Rewrite $p^{9}$ as ${(p^{3})}^{3}$ .

$1^{3} - {(p^{3})}^{3}$

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 $a = 1$ and $b = p^{3}$ .

$(1 - p^{3}) (1^{2} + 1 p^{3} + {(p^{3})}^{2})$

Simplify.

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

$(1^{3} - p^{3}) (1^{2} + 1 p^{3} + {(p^{3})}^{2})$

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 $a = 1$ and $b = p$ .

$(1 - p) (1^{2} + 1 p + p^{2}) (1^{2} + 1 p^{3} + {(p^{3})}^{2})$

Simplify.

One to any power is one.

$(1 - p) (1 + 1 p + p^{2}) (1^{2} + 1 p^{3} + {(p^{3})}^{2})$

Multiply $p$ by $1$ .

$(1 - p) (1 + p + p^{2}) (1^{2} + 1 p^{3} + {(p^{3})}^{2})$

Multiply $p^{3}$ by $1$ .

$(1 - p) (1 + p + p^{2}) (1^{2} + p^{3} + {(p^{3})}^{2})$

One to any power is one.

$(1 - p) (1 + p + p^{2}) (1 + p^{3} + {(p^{3})}^{2})$

Multiply the exponents in ${(p^{3})}^{2}$ .

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

$(1 - p) (1 + p + p^{2}) (1 + p^{3} + p^{3 \cdot 2})$

Multiply $3$ by $2$ .

$(1 - p) (1 + p + p^{2}) (1 + p^{3} + p^{6})$

Factor 1-p^9

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