How do you factor #1331x^3 – 8y^3#?

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Answer 1 · 2016-01-07T23:57:01

Use the difference of cubes identity to find:

#1331x^3-8y^3 =(11x-2y)(121x^2+22xy+4y^2)#

Both #1331x^3 = (11x)^3# and #8y^3 = (2y)^3# are perfect cubes.

Use the difference of cubes identity, which can be written:

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

with #a=11x# and #b=2y# as follows:

#1331x^3-8y^3#

#=(11x)^3-(2y)^3#

#=(11x-2y)((11x)^2+(11x)(2y)+(2y)^2)#

#=(11x-2y)(121x^2+22xy+4y^2)#

The remaining quadratic factor cannot be factored into linear factors with Real coefficients (as you can tell by checking its discriminant).

You can factor it using Complex coefficients:

#=(11x-2y)(11x-2omega y)(11x-2omega^2 y)#

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