How do you factor #a^9 + b^12 c^15#?

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Answer 1 · 2015-12-05T15:27:25

#a^9+b^12c^15 = (a^3+b^4c^5)(a^6-a^3b^4c^5+b^8c^10)#

#a^9+b^12c^15# is of the form #A^3+B^3# where #A=a^3# and #B=b^4c^5#.

The sum of cubes identity tells us that:

#A^3+B^3 = (A+B)(A^2-AB+B^2)#

Hence:

#a^9+b^12c^15=(a^3)^3+(b^4c^5)^3#

#= (a^3+b^4c^5)((a^3)^2-(a^3)(b^4c^5)+(b^4c^5)^2)#

#= (a^3+b^4c^5)(a^6-a^3b^4c^5+b^8c^10)#

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

#= (a^3+b^4c^5)(a^3+omega b^4c^5)(a^3+omega^2 b^4c^5)#

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