If \(a+b+c=0\) then the value of \[ \frac{a^2}{a^2-b c}+\frac{b^2}{b^2-c a}+\frac{c^2}{c^2-a b} \]

Solution

Given:
\[ a + b + c = 0 \]

We need to find the value of:
\[ \frac{a^2}{a^2 – bc} + \frac{b^2}{b^2 – ca} + \frac{c^2}{c^2 – ab} \]

Since \(a + b + c = 0\), we can write \(a = -(b + c)\).

Step 1: Substitute \(a = -(b + c)\)

\[ \frac{(-b – c)^2}{(-b – c)^2 – bc} + \frac{b^2}{b^2 – c(-b – c)} + \frac{c^2}{c^2 – b(-b – c)} \]

Step 2: Simplify

\[ \frac{(b + c)^2}{(b + c)^2 – bc} + \frac{b^2}{b^2 + bc – c^2} + \frac{c^2}{c^2 + bc – b^2} \]

Step 3: Simplify further

\[ \frac{(b + c)^2}{b^2 + c^2 + 2bc – bc} + \frac{b^2}{b^2 + c^2 + bc} + \frac{c^2}{b^2 + c^2 + bc} \]

\[ \frac{(b + c)^2}{b^2 + c^2 + bc} + \frac{b^2}{b^2 + c^2 + bc} + \frac{c^2}{b^2 + c^2 + bc} \]

Step 4: Combine the fractions

\[ \frac{(b + c)^2 + b^2 + c^2}{b^2 + c^2 + bc} = \frac{b^2 + c^2 + 2bc + b^2 + c^2}{b^2 + c^2 + bc} \]
\[ = \frac{2b^2 + 2c^2 + 2bc}{b^2 + c^2 + bc} \]
\[ = 2 \frac{b^2 + c^2 + bc}{b^2 + c^2 + bc} \]
\[ = 2 \]

Conclusion

The value of \(\frac{a^2}{a^2 – bc} + \frac{b^2}{b^2 – ca} + \frac{c^2}{c^2 – ab}\) is 2.