Question

a+b+c=0 prove it
a^2bc+b^2ab+c^2ca=3​

Answer 1

Given,

(a+b+c) = 0

(a+b+c)^3 = 0

» a^3 +b^3 +c^3 -3abc = 0

» a^3 +b^3 +c^3 = 3abc

divide throughout by abc.

» a^2bc+b^2ac +c^2ab=3