Factor by Grouping a^3-m^3n^9

$a^{3} - m^{3} n^{9}$

Step 1

Rewrite $m^{3} n^{9}$ as ${(m n^{3})}^{3}$ .

$a^{3} - {(m n^{3})}^{3}$

Step 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 $a = a$ and $b = m n^{3}$ .

$(a - (m n^{3})) (a^{2} + a (m n^{3}) + {(m n^{3})}^{2})$

Step 3

Simplify.

Remove parentheses.

$(a - m n^{3}) (a^{2} + a (m n^{3}) + {(m n^{3})}^{2})$

Remove parentheses.

$(a - m n^{3}) (a^{2} + a m n^{3} + {(m n^{3})}^{2})$

Apply the product rule to $m n^{3}$ .

$(a - m n^{3}) (a^{2} + a m n^{3} + m^{2} {(n^{3})}^{2})$

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

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

$(a - m n^{3}) (a^{2} + a m n^{3} + m^{2} n^{3 \cdot 2})$

Multiply $3$ by $2$ .

$(a - m n^{3}) (a^{2} + a m n^{3} + m^{2} n^{6})$

Factor by Grouping a^3-m^3n^9

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