Question : If $a+b+c+d=4$, then find the value of $\frac{1}{(1-a)(1-b)(1-c)}+\frac{1}{(1-b)(1-c)(1-d)}+\frac{1}{(1-c)(1-d)(1-a)}+\frac{1}{(1-d)(1-a)(1-b)}$?
Option 1: 0
Option 2: 5
Option 3: 1
Option 4: 4
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Correct Answer: 0
Solution :
Given: $a+b+c+d=4$
$\frac{1}{(1-a)(1-b)(1-c)}+\frac{1}{(1-b)(1-c)(1-d)}+\frac{1}{(1-c)(1-d)(1-a)}+\frac{1}{(1-d)(1-a)(1-b)}$
= $\frac{1-d+1-a+1-b+1-c}{(1-a)(1-b)(1-c)(1-d)}$
= $\frac{4-(a+b+c+d)}{(1-a)(1-b)(1-c)(1-d)}$
= $\frac{4-4}{(1-a)(1-b)(1-c)(1-d)}$
= $\frac{0}{(1-a)(1-b)(1-c)(1-d)}$
= $0$
Hence, the correct answer is 0.
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