Question 282889
128m^6 - 2n^6


2(64m^6 - n^6)


2((4m^2)^3 - (n^2)^3) 



2(4m^2 - n^2)((4m^2)^2+4m^2n^2+(n^2)^2)


2(4m^2 - n^2)(16m^4+4m^2n^2+n^4)



2(2m+n)(2m-n)(16m^4+4m^2n^2+n^4)



Answer: 128m^6 - 2n^6 = 2(2m+n)(2m-n)(16m^4+4m^2n^2+n^4)