Question : If $a+b+c=0$, then the value of $\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}$ is:
Option 1: 1
Option 2: 3
Option 3: - 1
Option 4: 0
Correct Answer: 3
Solution : $a+b+c=0$ $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ ⇒ $a^3+b^3+c^3 - 3abc = 0$ ⇒ $a^3+b^3+c^3 = 3abc$ ⇒ $\frac{a^3+b^3+c^3}{abc} = 3$ ⇒ $\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab} = 3$ Hence, the correct answer is 3.
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Question : If $a, b, c$ are all non-zero and $a+b+c=0$, then find the value of $\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{ab}$.
Option 1: $3$
Option 2: $4$
Option 3: $1$
Option 4: $\frac{1}{2}$
Question : If $(a^2 = b + c)$, $(b^2 = a + c)$, $(c^2 = b + a)$. Then, what will be the value of $(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})$?
Option 1: –1
Option 2: 2
Option 3: 1
Question : If $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}(a\neq b\neq c)$, then the value of $abc$ is:
Option 1: $\pm 1$
Option 2: $\pm 2$
Option 3: $0$
Option 4: $\pm \frac{1}{2}$
Question : If $a+b+c=0$, then the value of $\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}$ is:
Option 1: 2
Option 2: –2
Option 3: 0
Option 4: 4
Question : If $\frac{A}{L}+\frac{M}{B}=1$ and $\frac{B}{M}+\frac{N}{C}=1$, then the value of $\frac{L}{A}+\frac{C}{N}$ is:
Option 1: $\frac{B}{M}$
Option 2: 0
Option 4: $\frac{M}{B}$
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