Question

If a+b+c=0 what is the value of $\frac{a^2+b^2+ab}{b^2+c^2+bc}+\frac{c^2+ca+a^2}{b^2+c^2+bc}$ ?

Answer 1

Given, $a+b+c=0-eq.1\Rightarrow (a+b)=-c-eq.2\Rightarrow (a+b{)}^2=c^2-eq.3$

Similarly we get, $(b+c{)}^2=a^2\text{and}(a+c{)}^2=b^2$

Now, $\frac{a^2+b^2+ab}{b^2+c^2+bc}+\frac{c^2+ca+a^2}{b^2+c^2+bc}=\frac{a^2+b^2+ab+c^2+ca+a^2}{b^2+c^2+bc}=\frac{2a^2+2b^2+2ab+2c^2+2ca+2a^2}{2b^2+2c^2+2bc}\left[\text{multiply denominator and numerator with 2}\right]=\frac{(a^2+2ab+b^2)+(a^2+b^2)+(c^2+2ca+a^2)+(c^2+a^2)}{(b^2+bc+c^2)+b^2+c^2}=\frac{(a+b{)}^2+(c+a{)}^2+(2a^2+b^2+c^2)}{(b+c{)}^2+b^2+c^2}=\frac{c^2+b^2+2a^2+b^2+c^2}{a^2+b^2+c^2}\left[\text{from eq.1 , eq.2 , .eq.3}\right]=\frac{2(a^2+b^2+c^2)}{a^2+b^2+c^2}=2$