# How do you factor #8x^6 y^9 - 64a^3 b^12#?

##### 1 Answer

#### Explanation:

Note that both terms are divisible by

#8x^6y^9 - 64a^3b^12 = 8(x^6y^9 - 8a^3b^12)#

Note that both

Also note that:

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

So we can use this "difference of cubes identity" with

#x^6y^9 - 8a^3b^12#

#= (x^2y^3)^3 - (2ab^4)^3#

#= (x^2y^3-2ab^4)((x^2y^3)^2+(x^2y^3)(2ab^4)+(2ab^4)^2)#

#= (x^2y^3-2ab^4)(x^4y^6+2x^2y^3ab^4+4a^2b^8)#

Neither of the remaining factors has simpler factorisations using real coefficients. It is possible to factor the second decic expression using complex coefficients, most naturally using the primitive complex cube root of

#omega = -1/2+sqrt(3)/2i#

Then:

#x^4y^6+2x^2y^3ab^4+4a^2b^8 = (x^2y^3-2omegaab^4)(x^2y^3-2omega^2ab^4)#