Question

2.a^2(b+c)^2+b^2(c+a)^2+c^2(a+b)^2+abc(a+b+c)+(a^2+b^2+c^2)(

Answer 1

原式=a^2(b+c)^2+b^2(c+a)^2+c^2(a+b)^2+abc(a+b+c)+a^3(b+c)+b^3(a+c)+c^3(a+b)+a^2bc+ab^2c+abc^2

=[a^2(b+c)^2+a^3(b+c)]+[b^2(a+c)^2+b^3(a+c)]+[c^2(a+b)+c^3(a+b)]+abc(a+b+c)+abc(a+b+c)

=a^2(b+c)(a+b+c)+b^2(a+c)(a+b+c)+c^2(a+b)(a+b+c)+2abc(a+b+c)

=(a+b+c)(ba^2+ca^2+b^2a+b^2c++c^2a+c^2b)

=(a+b+c)[(b+c)a^2+(b^2+2bc+c^2)a+bc(b+c)]

=(a+b+c)[(b+c)a^2+(b+c)^2+bc(b+c)]

=(a+b+c)(b+c)[a^2+(b+c)a+bc]

=(a+b+c)(b+c)(a+b)(a+c)