How do you factor #e^6 + f^3#?

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Answer 1 · 2015-12-09T19:03:21

Use the sum of cubes identity to find:

#e^6+f^3 = (e^2+f)(e^4-e^2f+f^2)#

The sum of cubes identity may be written:

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

If we let #a=e^2# and #b=f# then we find:

#e^6+f^3 = (e^2)^3+f^3#

#=(e^2+f)((e^2)^2-(e^2)f+f^2)#

#=(e^2+f)(e^4-e^2f+f^2)#

If we allow Complex coefficients then this can be factored further:

#=(e^2+f)(e^2+omega f)(e^2+omega^2 f)#

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