Factor 729a^6-64b^6

$729 a^{6} - 64 b^{6}$

Rewrite $729 a^{6}$ as ${(9 a^{2})}^{3}$ .

${(9 a^{2})}^{3} - 64 b^{6}$

Rewrite $64 b^{6}$ as ${(4 b^{2})}^{3}$ .

${(9 a^{2})}^{3} - {(4 b^{2})}^{3}$

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 $a = 9 a^{2}$ and $b = 4 b^{2}$ .

$(9 a^{2} - (4 b^{2})) ({(9 a^{2})}^{2} + 9 a^{2} (4 b^{2}) + {(4 b^{2})}^{2})$

Simplify.

Remove parentheses.

$(9 a^{2} - 1 \cdot 4 b^{2}) ({(9 a^{2})}^{2} + 9 a^{2} (4 b^{2}) + {(4 b^{2})}^{2})$

Rewrite $9 a^{2}$ as ${(3 a)}^{2}$ .

$({(3 a)}^{2} - 1 \cdot 4 b^{2}) ({(9 a^{2})}^{2} + 9 a^{2} (4 b^{2}) + {(4 b^{2})}^{2})$

Rewrite $4 b^{2}$ as ${(2 b)}^{2}$ .

$({(3 a)}^{2} - {(2 b)}^{2}) ({(9 a^{2})}^{2} + 9 a^{2} (4 b^{2}) + {(4 b^{2})}^{2})$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 a$ and $b = 2 b$ .

$(3 a + 2 b) (3 a - (2 b)) ({(9 a^{2})}^{2} + 9 a^{2} (4 b^{2}) + {(4 b^{2})}^{2})$

Multiply $2$ by $- 1$ .

$(3 a + 2 b) (3 a - 2 b) ({(9 a^{2})}^{2} + 9 a^{2} (4 b^{2}) + {(4 b^{2})}^{2})$

Multiply $4$ by $9$ .

$(3 a + 2 b) (3 a - 2 b) ({(9 a^{2})}^{2} + 36 a^{2} b^{2} + {(4 b^{2})}^{2})$

Apply the product rule to $9 a^{2}$ .

$(3 a + 2 b) (3 a - 2 b) (9^{2} {(a^{2})}^{2} + 36 a^{2} b^{2} + {(4 b^{2})}^{2})$

Raise $9$ to the power of $2$ .

$(3 a + 2 b) (3 a - 2 b) (81 {(a^{2})}^{2} + 36 a^{2} b^{2} + {(4 b^{2})}^{2})$

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

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

$(3 a + 2 b) (3 a - 2 b) (81 a^{2 \cdot 2} + 36 a^{2} b^{2} + {(4 b^{2})}^{2})$

Multiply $2$ by $2$ .

$(3 a + 2 b) (3 a - 2 b) (81 a^{4} + 36 a^{2} b^{2} + {(4 b^{2})}^{2})$

Apply the product rule to $4 b^{2}$ .

$(3 a + 2 b) (3 a - 2 b) (81 a^{4} + 36 a^{2} b^{2} + 4^{2} {(b^{2})}^{2})$

Raise $4$ to the power of $2$ .

$(3 a + 2 b) (3 a - 2 b) (81 a^{4} + 36 a^{2} b^{2} + 16 {(b^{2})}^{2})$

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

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

$(3 a + 2 b) (3 a - 2 b) (81 a^{4} + 36 a^{2} b^{2} + 16 b^{2 \cdot 2})$

Multiply $2$ by $2$ .

$(3 a + 2 b) (3 a - 2 b) (81 a^{4} + 36 a^{2} b^{2} + 16 b^{4})$

Factor.

Rewrite $81 a^{4} + 36 a^{2} b^{2} + 16 b^{4}$ in a factored form.

Rewrite the middle term.

$(3 a + 2 b) (3 a - 2 b) (81 a^{4} + 2 (9 a^{2}) (4 b^{2}) - 36 a^{2} b^{2} + 16 b^{4})$

Rearrange terms.

$(3 a + 2 b) (3 a - 2 b) (81 a^{4} + 2 (9 a^{2}) (4 b^{2}) + 16 b^{4} - 36 a^{2} b^{2})$

Factor first three terms by perfect square rule.

$(3 a + 2 b) (3 a - 2 b) ({(9 a^{2} + 4 b^{2})}^{2} - 36 a^{2} b^{2})$

Rewrite $36 a^{2} b^{2}$ as ${(6 a b)}^{2}$ .

$(3 a + 2 b) (3 a - 2 b) ({(9 a^{2} + 4 b^{2})}^{2} - {(6 a b)}^{2})$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 9 a^{2} + 4 b^{2}$ and $b = 6 a b$ .

$(3 a + 2 b) (3 a - 2 b) ((9 a^{2} + 4 b^{2} + 6 a b) (9 a^{2} + 4 b^{2} - (6 a b)))$

Simplify.

Multiply $6$ by $- 1$ .

$(3 a + 2 b) (3 a - 2 b) ((9 a^{2} + 4 b^{2} + 6 a b) (9 a^{2} + 4 b^{2} - 6 (a b)))$

Remove parentheses.

$(3 a + 2 b) (3 a - 2 b) ((9 a^{2} + 4 b^{2} + 6 a b) (9 a^{2} + 4 b^{2} - 6 a b))$

Remove unnecessary parentheses.

$(3 a + 2 b) (3 a - 2 b) (9 a^{2} + 4 b^{2} + 6 a b) (9 a^{2} + 4 b^{2} - 6 a b)$

Factor 729a^6-64b^6

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