Question 56
If a + b + c = 0, then the value of $$\frac{a^{2} + b^{2} + c^{2}}{a^{2} - bc}$$ is:
Solution
$$a+b+c=0$$
$$a\ =\ -b-c$$
Squaring both sides:
$$a^2=b^2+c^2+2bc$$.
Adding $$b^2+c^2$$ on both sides, we get:
$$a^2+b^2+c^2=2\left(b^2+c^2+bc\right)$$.
Substituting values of $$a^2+b^2+c^2\ and\ a^2$$ in the given expression, we get:
$$\frac{2\left(b^2+c^2+bc\right)}{b^2+c^2+2bc-bc}=2$$.
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