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

##### 1 Answer

Dec 5, 2015

#### Explanation:

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