2023/025) GIven $a+b+c = 0$ Find the value of $\frac{ab}{a^2+ab+b^2} + \frac{bc}{b^2 + bc+c^2} + \frac{ca}{c^2 + ca + b^2}$

We have $a+b+c = 0$

Hence $ a+ b = -c $

Squaring both sides $a^2+2ab + b^2 = c^2$

Adding  $a^2+b^2$ on both sides we get

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

Or $a^2+ab + b^2 = \frac{1}{2}(a^2+b^2+c^2)$

So $\frac{ab}{a^2+ab + b^2} = \frac{2ab}{a^2+b^2+c^2}\cdots(1)$

Similarly we have  $\frac{bc}{b^2+bc + c^2} = \frac{2bc}{a^2+b^2+c^2}\cdots(2)$

And $\frac{ca}{c^2+ca + a^2} = \frac{2ca}{a^2+b^2+c^2}\cdots(3)$

Adding (1) (2) and (3) we get

$\frac{ab}{a^2+ab+b^2} + \frac{bc}{b^2 + bc+c^2} + \frac{ca}{c^2 + ca + b^2}= \frac{2ab+2bc+2ca}{a^2+b^2+c^2}\cdots(4)$

Now staring with $a+b+c=0$ squaring both sides we get

$a^2+b^2+c^2 + 2ab + 2bc+2ca= 0$

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

Or $\frac{2ab+2bc+2ca}{a^2 +b^2+ c^2} = -1\cdots(5)$

Form (4) and (5) we get $\frac{ab}{a^2+ab+b^2} + \frac{bc}{b^2 + bc+c^2} + \frac{ca}{c^2 + ca + b^2}= - 1$