Question

已知a^3+b^3+c^3-3abc=3(a+b+c),则(a-b)^2+(b-c)^2+(a-b)(b-c)=?

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)

即：(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=3(a+b+c)

又：a+b+c≠0

所以：a^2+b^2+c^2-ab-ac-bc=3

(a-b)^2+(b-c)^2+(a-b)(b-c)

=a^2-2ab+b^2+b^2-2bc+c^2+ab-ac-b^2+bc

=a^2+b^2+c^2-ab-bc-ac

=3