Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Let $a, b$ and $c$ be such that $a+b+c=0$ and $P =\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}$ is defined. What is the value of $P$? $Solution:$ $2a^2+bc = a^2 + a^2 -(a+c)c$, since $b=-(a+c)$. $\implies 2a^2+bc = a^2 + a^2 -ac-c^2 = (a-c)(a+c)+a(a-c) = (a-c)(2a+c)=(a-c)(2a-a-b)=(a-c)(a-b)$ Therefore $\frac{a^2}{2a^2+bc} = \frac{a^2}{(a-c)(a-b)}$ Similarly, $\frac{b^2}{2b^2+ac} = \frac{b^2}{(b-a)(b-c)}$ and $\frac{c^2}{2c^2+ab} = \frac{c^2}{(c-a)(c-b)}$ Therefore $P = \frac{a^2}{(a-c)(a-b)} +\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)} = \frac{1}{a-b} \big( \frac{a^2}{a-c} - \frac{b^2}{b-c} \big)+\frac{c^2}{(c-a)(c-b)}$ $\implies P = \frac{ab-ca-cb}{(c-a)(c-b)}+ \frac{c^2}{(c-a)(c-b)} = \frac{ab-ca-cb+c^2}{(c-a)(c-b)} = \frac{(c-a)(c-b)}{(c-a)(c-b)} = 1$