Factor 1+q^9

$1 + q^{9}$

Step 1

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

$1^{3} + q^{9}$

Step 2

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

$1^{3} + {(q^{3})}^{3}$

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

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

Step 4

Simplify.

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

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

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 $a = 1$ and $b = q$ .

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

Simplify.

One to any power is one.

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

Rewrite $- 1 q$ as $- q$ .

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

One to any power is one.

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

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

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

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

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

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

Multiply $3$ by $2$ .

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

Factor 1+q^9

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