How do you factor #x^3y^6 + 8#?

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Answer 1 · 2015-12-14T07:17:12

#x^3y^6 + 8 = (xy^2+2)(x^2y^4-2xy^2+4)#

The sum of cubes identity can be written:

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

Both #x^3y^6 = (xy^2)^3# and #8 = 2^3# are perfect cubes, so let #a = xy^2# and #b=2# to find:

#x^3y^6+8#

#=(xy^2)^2+2^3#

#=(xy^2+2)((xy^2)^2-(xy^2)(2)+2^2)#

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

That's as far as you can get with Real coefficients, but if you allow Complex coefficients this factors a little further:

#=(xy^2+2)(xy^2+2omega)(xy^2+2omega^2)#

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