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);

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

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

## No comments:

## Post a Comment