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#.