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 3: 0
Option 4: 1
Correct Answer: 2
Solution : Given: $a^{2}+b^{2}+c^{2}=ab+bc+ca$ ⇒ $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$ ⇒ $\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0$ ⇒ $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$ The square of any real number is always non-negative. Therefore, we have: $(a-b)^{2}=0 \Rightarrow a=b$ $(b-c)^{2}=0 \Rightarrow b=c$ $(c-a)^{2}=0 \Rightarrow c=a$ Thus, we have $a=b=c$. Now, we can calculate $\frac{a+c}{b}$ as follows: $\frac{a+c}{b}=\frac{a+a}{a}=2$ Hence, the correct answer is 2.
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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 4: 4
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
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 3: 1
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