How do you factor #((x-y)^3) +8#?

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Answer 1 · 2016-08-18T19:55:57

#(x-y)^3+8#

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

#=(x-y+2)(x-y-1+sqrt(3)i)(x-y-1-sqrt(3)i)#

The sum of cubes identity can be written:

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

Use this with #a=(x-y)# and #b=2# to find:

#(x-y)^3+8#

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

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

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

If you allow Complex coefficients, then this factors further as:

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

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

#=(x-y+2)(x-y-1+sqrt(3)i)(x-y-1-sqrt(3)i)#