How do you factor $2 y^{3} - 128$ $2y^3 -128$ ?

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How do you factor $2 y^{3} - 128$ $2y^3 -128$ ?

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Answer 1 · 2016-05-29T00:11:51

$2 y^{3} - 128 = 2 (y - 4) (y^{2} + 4 y + 16)$ $2y^3-128=2(y-4)(y^2+4y+16)$

Explanation:

Separate out the common scalar factor $2$ $2$ , then factor as a difference of cubes:

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

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

$2 y^{3} - 128$ $2y^3-128$

$= 2 (y^{3} - 64)$ $=2(y^3-64)$

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

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

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

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How do you factor $2 y^{3} - 128$ $2y^3 -128$ ?

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