Question : If $a+b+c=3$ and none of $a,b$ and $c$ is equal to $1$, what is the value of $\frac{1}{(1-a)(1-b)}+\frac{1}{(1-b)(1-c)}+\frac{1}{(1-c)(1-a)}?$
Option 1: 0
Option 2: 1
Option 3: 3
Option 4: 6
Correct Answer: 0
Solution : Given: $a+b+c=3$ and none of $a, b$ and $c$ is equal to 1. $\frac{1}{(1-a)(1-b)}+\frac{1}{(1-b)(1-c)}+\frac{1}{(1-c)(1-a)}$ = $\frac{1-c+1-b+1-c}{(1-a)(1-b)(1-c)}$ = $\frac{3-(a+b+c)}{(1-a)(1-b)(1-c)}$ = $\frac{3-3}{(1-a)(1-b)(1-c)}$ = $\frac{0}{(1-a)(1-b)(1-c)}$ = $0$ Hence, the correct answer is 0.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
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:
Question : If $a, b, c$ are all non-zero and $a+b+c=0$, then find the value of $\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{ab}$.
Question : If $(a^2 = b + c)$, $(b^2 = a + c)$, $(c^2 = b + a)$. Then, what will be the value of $(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})$?
Question : If $a^{2}+b^{2}+c^{2}=ab+bc+ca,$ then the value of $\frac{a+c}{b}$ is:
Question : If $\left (2a-3 \right )^{2}+\left (3b+4 \right )^{2}+\left ( 6c+1\right)^{2}=0$, then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}+3$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile