How do you factor $27 y^{3} + 64$ $27y^3+64$ ?

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How do you factor $27 y^{3} + 64$ $27y^3+64$ ?

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Answer 1 · 2016-05-02T11:42:32

$27 y^{3} + 64 = (3 y + 4) (9 y^{2} - 12 y + 16)$ $27y^3+64=(3y+4)(9y^2-12y+16)$

Explanation:

Note that both $27 y^{3} = {(3 y)}^{3}$ $27y^3 = (3y)^3$ and $64 = 4^{3}$ $64=4^3$ are perfect cubes.

So it is natural to use the sum of cubes identity:

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

with $a = 3 y$ $a=3y$ and $b = 4$ $b=4$ as follows:

$27 y^{3} + 64$ $27y^3+64$

$= {(3 y)}^{3} + 4^{3}$ $=(3y)^3+4^3$

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

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

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How do you factor $27 y^{3} + 64$ $27y^3+64$ ?

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