If (ab+bc+ca)=0. then find value of (1\a^2-bc + 1\b^2-ca +1\c^2-ab)
Take -bc from ab+bc+ca u get= a(b+c)=-bc put value of bc in first u get 1\a(a+b+c) in first nd as in next term then take 1\(a+b+c) common from all three term then u get (1\a+1\b+1\c) solve it (ab+bc+ca)\abc which is zero ###if any query write again###
I have tried it by supposition taking a=b=2 and c=-1. And it isn't coming 0. Do we have options- vipin?