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.