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.
Related Questions
Know More about
Staff Selection Commission Sub Inspector ...
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Get Updates Brochure
Your Staff Selection Commission Sub Inspector Exam brochure has been successfully mailed to your registered email id “”.
