Question : If $a+b+c=0$, the value of $\frac{a^2+b^2+c^2}{a^2-bc}$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: 3
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Correct Answer: 2
Solution : Given: $a+b+c=0$ $\therefore b+c=-a$ Now, $a+b+c=0$ By squaring both sides, we get, $(a+b+c)^2=0$ ⇒ $a^2+b^2+c^2+2(ab+bc+ca)=0$ ⇒ $a^2+b^2+c^2+2[a(b+c)+bc]=0$ ⇒ $a^2+b^2+c^2+2[a(-a)+bc]=0$ ⇒ $a^2+b^2+c^2-2[a^2-bc]=0$ ⇒ $a^2+b^2+c^2=2[a^2-bc]$ $\therefore\frac{a^2+b^2+c^2}{a^2-bc}=2$ Hence, the correct answer is $2$.
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Question : If $a+b+c = 0$, then the value of $\small \frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{(c+a)(a+b)}$ is:
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:
Question : If $\frac{2+a}{a}+\frac{2+b}{b}+\frac{2+c}{c}=4$, then the value of $\frac{ab+bc+ca}{abc}$ is:
Question : If $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}$ where $a \neq b\neq c\neq 0$, then the value of $a^{2}b^{2}c^{2}$ is:
Question : If $\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$, then find the value of $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$.
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