Question

推导a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-3abc=(a+b+c)^3

Answer 1

a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-3abc

=a³+b³+c³+3a²b+3abc+3a²c+3ab²+3b²c+3abc+3abc+3bc²+3ac²-3abc

=(a+b)³+c³+3a²c+3ac²+3b²c+3bc²+3abc+3abc

=(a+b+c)(a²+2ab+b²-ac-bc+c²)+3ac(a+b+c)+3bc(a+b+c)

=(a+b+c)(a²+2ab+b²-ac-bc+3ac+3bc+c²)

=(a+b+c)(a²+2ab+b²+2ac+2bc+c²)

=(a+b+c)(a+b+c)²

=(a+b+c)³