Factor y^3-64z^3

$y^{3} - 64 z^{3}$

Step 1

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

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

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

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

Step 3

Simplify.

Multiply $4$ by $- 1$ .

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

Rewrite using the commutative property of multiplication.

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

Apply the product rule to $4 z$ .

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

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

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

Factor y^3-64z^3

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