Factor 27a^4b^3+64a^7

$27 a^{4} b^{3} + 64 a^{7}$

Step 1

Factor $a^{4}$ out of $27 a^{4} b^{3} + 64 a^{7}$ .

Factor $a^{4}$ out of $27 a^{4} b^{3}$ .

$a^{4} (27 b^{3}) + 64 a^{7}$

Factor $a^{4}$ out of $64 a^{7}$ .

$a^{4} (27 b^{3}) + a^{4} (64 a^{3})$

Factor $a^{4}$ out of $a^{4} (27 b^{3}) + a^{4} (64 a^{3})$ .

$a^{4} (27 b^{3} + 64 a^{3})$

Step 2

Rewrite $27 b^{3}$ as ${(3 b)}^{3}$ .

$a^{4} ({(3 b)}^{3} + 64 a^{3})$

Step 3

Rewrite $64 a^{3}$ as ${(4 a)}^{3}$ .

$a^{4} ({(3 b)}^{3} + {(4 a)}^{3})$

Step 4

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 $a = 3 b$ and $b = 4 a$ .

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

Step 5

Factor.

Simplify.

Apply the product rule to $3 b$ .

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

Raise $3$ to the power of $2$ .

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

Multiply $3$ by $- 1$ .

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

Rewrite using the commutative property of multiplication.

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

Multiply $- 3$ by $4$ .

$a^{4} ((3 b + 4 a) (9 b^{2} - 12 b a + {(4 a)}^{2}))$

Apply the product rule to $4 a$ .

$a^{4} ((3 b + 4 a) (9 b^{2} - 12 b a + 4^{2} a^{2}))$

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

$a^{4} ((3 b + 4 a) (9 b^{2} - 12 b a + 16 a^{2}))$

Remove unnecessary parentheses.

$a^{4} (3 b + 4 a) (9 b^{2} - 12 b a + 16 a^{2})$

Factor 27a^4b^3+64a^7

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