2014/094) If $a+b+c=0$ then prove that $\dfrac{a^2}{a^2-bc} +\dfrac{b^2}{b^2-ca} +\dfrac{c^2}{c^2-ab} =2$

Solution

We are given $a + b+ c = 0\cdots(1)$

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

 so 

$a^2 +b^2 + c^2 = -2(ab +bc +ca)\cdots(2)$

 From (1)
$a(a+b+c) = 0$

$a^2 + ab + ca = 0$

or $a^2 - bc = -(ab + bc+ ca)\cdots(3)$

Similarly  

$b^2 - ca = -(ab + bc+ ca)\cdots(4)$

$c^2 - ab = -(ab + bc+ ca)\cdots(5)$

Thus, using (3), (4) & (5), we can rearrange the LHS as follows;

LHS = $\dfrac{a^2}{a^2-bc} +\dfrac{b^2}{b^2-ca} +\dfrac{c^2}{c^2-ab}$

 = $\dfrac{a^2 +b^2+ c^2}{-ab-bc-ca}$

= -$\dfrac{a^2 +b^2+ c^2}{ab+bc+ca}$

= -(- 2)$\dfrac{ab+bc+ca}{ab+bc+ca}$

= 2