一道好难的证明题！你们一定要帮我！

a^3+b^3+c^3-3abc
=(a^3+3a^2b+3ab^2+b^3+c^3)-(3abc+3a^2b+3ab^2)
=[(a+b)^3+c^3]-3ab(a+b+c)
=(a+b+c)(a^2+b^2+2ab-ac-bc+c^2)-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2+2ab-3ab-ac-bc)
=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
所以a^3+b^3+c^3
=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc

所以左边=(a+b+c)^3+2(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+6abc-3(a^2+b^2+c^2)(a+b+c)
=(a+b+c)[(a+b+c)^2+2(a^2+b^2+c^2-ab-bc-ac)-3(a^2+b^2+c^2)]+6abc
=(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ca+2a^2+2b^2+2c^2-2ab-2bc-2ac-3a^2-3b^2-3c^2)+6abc
=(a+b+c)*0+6abc
=6abc=右边

a^3+b^3+c^3-3abc
=[( a+b)^3-3a^2b-3ab^2]+c^3-3abc
=[(a+b)^3+c^3]-(3a^2b+3ab^2+3abc)
=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)
=(a+b+c)(a^2+b^2+2ab-ac-bc+c^2)-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
=1/2{(a+b+c)[-(a^2+b^2+c^2+2ab+2bc+2ac)+3(a^2+b^2+c^2)]}
即:
2(a^3+b^3+c^3)-6abc=(a+b+c)[-(a+b+c)^2+3(a^2+b^2+c^2)]
展开得:
2(a^3+b^3+c^3)-6abc=-(a+b+c)^3+3(a+b+c)(a^2+b^2+c^2)<