If a+b+c=0 then prove that a2/a2-bc + b2/b2-ca + c2/c2-ab = 2

 If (a+b+c) = 0 --- --- --- --- --- --- --- --- (1); 

Multiply eqn (1) by a; We get, 

a(a+b+c) = 0  that is,  a^2 + ab + ca = 0  that is, 

a^2 - bc = -(ab + bc+ ca) --- --- --- --- --- --- --- --- (7); Similarly, multiplying eqn (1) by b, we can get, 

b^2 - ca = -(ab + bc+ ca) --- --- --- --- --- --- --- --- (8);  And again, multiplying eqn (1) by c, we can get, 

c^2 - ab = -(ab + bc+ ca) --- --- --- --- --- --- --- --- (9);  

Thus, using (7), (8) & (9), we can rearrange the LHS as follows; 

LHS = a^2/(a^2-bc) + b^2/(b^2-ca) + c^2/(c^2-ab) = (a^2 +b^2+ c^2)/(-ab-bc-ca) --- --- --- --- (10);  

From (4), (5) & (10), we get, 

LHS = p/q --- --- --- --- --- --- --- --- (11); 

And, from (6) & (11), we get, LHS = 2  that is, 

a^2/(a^2-bc) + b^2/(b^2-ca) + c^2/(c^2-ab) = 2