Ta có: a3+b3+c3-3abc=(a+b)3-3ab(a+b)+c3-3abc
=\(\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b\right)-3abc\)
=(a+b+c) \(\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
=(a+b+c)(a2+2ab+b2-ac-bc+c2-3ab)
=(a+b+c)(a2+b2+c2-ab-ac-bc)