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Algebra
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If a + b + c = 3 and none of a, b and c is equal to 1, then what is the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - a} \right)}}?$$

Answer & Solution
Correct Answer: Option A
$$\eqalign{ & a + b + c = 3 \cr & {\text{Put }}a = 4,\,b = - 1,\,c = 0 \cr & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - a} \right)}} \cr & = \frac{1}{{ - 3 \times 2}} + \frac{1}{{2 \times 1}} - \frac{1}{3} \cr & = \frac{1}{2} - \frac{1}{3} - \frac{1}{6} \cr & = \frac{0}{6} \cr & = 0 \cr} $$
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