If ab + bc + ca = 0, find the value of 1/a2 – bc + 1/b2 – ca + 1/c2-ab

(a2 means a square. I could not put it up)

Dear student,

given,ab+bc+ca=0-(1)
squaring both sides we get, (ab+bc+ca)^2 = 0
=>a^2b^2+b^2c^2+c^2a^2=-2ab^2c - 2abc^2 - 2a^2bc -(2)
now probelm: 1/(a^2 -bc) +1/(b^2 -ca) +1/(c^2- ab)
= (b^2 -ca) *(c^2 -ab) + (c^2 -ab)*(a^2- bc)+(a^2-bc)*(b^2 -ca) / (a^2 -bc) *(b^2 -ca) (c^2-ab)
solving and simplifying we will get
= - (a^2+ b^2 +c^2) * (ac+ab+bc) / (a^2 -bc) *(b^2- ca) *(c^2 -ab)
=0 (from eqn 1)
so, 1/a2 – bc + 1/b2 – ca + 1/c2-ab =0 [ans]

Regards, Soumya Ranjan Mohanty.