一道关于因式分解的题啊急用！

A^2+B^2+C^2-AB-BC-CA
=0.5*2(A^2+B^2+C^2-AB-BC-CA)
=0.5(A^2-2AB+B^2+B^2-2BC+C^2+C^2-2CA+A^2)
=0.5[(A-B)^2+(B-C)^2+(C-A)^2]
=0.5[(2002X+2003-2002X-2004)^2+(2002X+2004-2002X-2005)^2+(2002X+2005-2002X-2003)^2]
=0.5[(-1)^2+(-1)^2+2^2]
=0.5*6=3

a^2+b^2+c^2-ab-ac-bc
=(a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2)/2
=[(a-b)^2+(a-c)^2+(b-c)^2]/2
=[(2002x+2003-2002x-2004)^2+(2002x+2003-2002x-2005)^2+(2002x+2004-2002x-2005)^2]/2
=[(-1)^2+(-2)^2+(-1)^2]/2
=(1+4+1)/2=6/2=3