Factor 64-27y^3

$64 - 27 y^{3}$

Step 1

Rewrite $64$ as $4^{3}$ .

$4^{3} - 27 y^{3}$

Step 2

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

$4^{3} - {(3 y)}^{3}$

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

$(4 - (3 y)) (4^{2} + 4 (3 y) + {(3 y)}^{2})$

Step 4

Simplify.

Multiply $3$ by $- 1$ .

$(4 - 3 y) (4^{2} + 4 (3 y) + {(3 y)}^{2})$

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

$(4 - 3 y) (16 + 4 (3 y) + {(3 y)}^{2})$

Multiply $3$ by $4$ .

$(4 - 3 y) (16 + 12 y + {(3 y)}^{2})$

Apply the product rule to $3 y$ .

$(4 - 3 y) (16 + 12 y + 3^{2} y^{2})$

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

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

Factor 64-27y^3

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