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
Correct Answer: –2
Solution : Given: $a+b+c=0$ We know that $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$ Putting the value of $a+b+c=0$, we have, ⇒ $(0)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$ ⇒ $a^{2}+b^{2}+c^{2}=–2(ab+bc+ca)$ ⇒ $\frac{(a^{2}+b^{2}+c^{2})}{(ab+bc+ca)}=–2$ Hence, the correct answer is –2.
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Question : If $a^{2}+b^{2}+c^{2}=ab+bc+ca,$ then the value of $\frac{a+c}{b}$ is:
Option 1: 3
Option 2: 2
Option 4: 1
Question : The numerical value of $\frac{(a–b)^{2}}{(b–c)(c–a)}+\frac{(b–c)^{2}}{(c–a)(a–b)}+\frac{(c–a)^{2}}{(a–b)(b–c)}$ is: $(a\neq b\neq c)$
Option 1: $0$
Option 2: $1$
Option 3: $\frac{1}{3}$
Option 4: $3$
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 $a+b+c=0$, 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
Question : If $a+b+c=1, ab+bc+ca=-1$ and $abc=-1$, then the value of $a^{3}+b^{3}+c^{3}$ is:
Option 2: – 1
Option 3: 2
Option 4: – 2
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