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

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Answer 1 · 2016-08-02T22:38:52

$w^3-2y^3=(w-root(3)(2)y)(w^2+root(3)(2)wy+root(3)(4)y^2)$

The difference of cubes identity can be written:

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

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If the coefficient of $y^3$ was a perfect cube then the example would factor naturally as a difference of two cubes.

As it is, we need to use irrational coefficients to make it into a difference of cubes:

$w^3-2y^3$

$=w^3-(root(3)(2)y)^3$

$=(w-root(3)(2)y)(w^2+w(root(3)(2)y)+(root(3)(2)y)^2)$

$=(w-root(3)(2)y)(w^2+root(3)(2)wy+root(3)(4)y^2)$

The remaining quadratic factor can only be factored further with Complex coefficients, mentioned here for completeness:

$w^2+root(3)(2)wy+root(3)(4)y^2=(w-omega root(3)(2)y)(w-omega^2 root(3)(2)y)$

where $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$ .

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