a^3+b^3+c^3-3abc
=(a+b)^3-3ab(a+b)+c^3-3abc
=(a+b)^3+c^3-3ab(a+b)-3abc
=(a+b+c)^3-3(a+b)*c*(a+bc)-3ab(a+b+c)
=(a+b+c)*[(a+b+c)^2-3ab]
=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
(a^3+b^3+c^3-3abc)/(a+b+c)=a²+b²+c²-ab-ac-bc=3
(a-b)²+(b-c)²+(a-b)(b-c)
=a²-2ab+b²+b²-2bc+c²+ab-b²-ac+bc
=a²+b²+c²-ab-ac-bc
=3