Question : If $\frac{a^{2} - bc}{a^{2}+bc}+\frac{b^{2}-ca}{b^{2}+ca}+\frac{c^{2}-ab}{c^{2}+ab}=1$, then the value of $\frac{a^{2}}{a^{2}+bc}+\frac{b^{2}}{b^{2}+ac}+\frac{c^{2}}{c^{2}+ab}$ is:
Option 1: 0
Option 2: 1
Option 3: –1
Option 4: 2
Correct Answer: 2
Solution :
$\frac{a^{2}-bc}{a^{2}+bc}+\frac{b^{2}-ca}{b^{2}+ca}+\frac{c^{2}-ab}{c^{2}+ab}=1$
Adding 3 on both sides,
$⇒(\frac{a^{2}-bc}{a^{2}+bc}+1)+(\frac{b^{2}-ca}{b^{2}+ca}+1)+(\frac{c^{2}-ab}{c^{2}+ab}+1)=1+3$
$⇒(\frac{a^{2}-bc+a^{2}+bc}{a^{2}+bc})+(\frac{b^{2}-ca+b^{2}+ca}{b^{2}+ca})+(\frac{c^{2}-ab+c^{2}+ab}{c^{2}+ab})=4$
$⇒(\frac{2a^{2}}{a^{2}+bc})+(\frac{2b^{2}}{b^{2}+ca})+(\frac{2c^{2}}{c^{2}+ab})=4$
$⇒\frac{a^{2}}{a^{2}+bc}+\frac{b^{2}}{b^{2}+ac}+\frac{c^{2}}{c^{2}+ab}=2$
Hence, the correct answer is 2.
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