Factor 27a^3-64b^3

$27 a^{3} - 64 b^{3}$

Step 1

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

${(3 a)}^{3} - 64 b^{3}$

Step 2

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

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

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

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

Step 4

Simplify.

Multiply $4$ by $- 1$ .

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

Apply the product rule to $3 a$ .

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

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

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

Rewrite using the commutative property of multiplication.

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

Multiply $3$ by $4$ .

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

Apply the product rule to $4 b$ .

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

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

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

Factor 27a^3-64b^3

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