Factor 64+y^3 - MathMaster

$64 + y^{3}$

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

$4^{3} + y^{3}$

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

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

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

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

Factor 64+y^3

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