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

Question

3215 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-29T00:11:51

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

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

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

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

$2y^3-128$

$=2(y^3-64)$

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

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

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

How do you factor 2y^3 -128 2y^3 -128?