My Friend's Failure 01

We know, $(a+b+c)^{3} = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)$

Therefore, $(a+b+c)^{3} = a^3 + b^3 + c^3$ will be true, if $(a+b)(b+c)(c+a) = 0$ $\Rightarrow (a+b)(bc+ab+c^2+ca) = 0$ $\Rightarrow abc+a^2b+c^2a+ca^2+b^2c+ab^2+bc^2+abc = 0$ $\Rightarrow a^2b+abc+ca^2+ab^2+b^2c+abc+abc+bc^2+c^2a = abc$ $\Rightarrow (a+b+c)(ab+bc+ca) = abc$

So, the answer is $\boxed {(a+b+c)(ab+bc+ca) = abc}$