How do you factor $1/8x^3-1/27y^3$ ?

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Answer 1 · 2015-05-21T21:52:05

$1/8x^3 - 1/27y^3$

$= (1/2)^3x^3-(1/3)^3y^3$

$= (1/2x)^3-(1/3y)^3$

$= (x/2)^3-(y/3)^3$

This is of the form $(a^3-b^3)$ which has a well known factorization:

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

So we can substitute $a = x/2$ , $b = y/3$ to get

$(x/2)^3-(y/3)^3$

$= (x/2-y/3)((x/2)^2+(x/2)(y/3)+(y/3)^2)$

$= (x/2-y/3)(x^2/4+(xy)/6+y^2/9)$

This is as far as we can go with real valued coefficients.

How do you factor 1/8x^3-1/27y^31/8x^3-1/27y^3?