Question

a^3+b^3+c^3-3ab=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=1/2(a+b+c)[(a-

Answer 1

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)

二个公式：

a^3+b^3=(a+b)(a^2-ab+b^2)

(a+b)^3=a^3+b^3+3a^2b+3ab^2