If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

Solution(By Examveda Team)

$$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & {\text{Put }} \cr & a = 0 \cr & b = 0 \cr & c = 2 \cr & d = 2 \cr & \therefore a + b + c + d = 4 \cr & \Rightarrow 0 + 0 + 2 + 2 = 4 \cr & \Rightarrow 4 = 4\left( {{\text{ satisfy}}} \right) \cr & \frac{1}{{\left( {1 - 0} \right)\left( {1 - 0} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 0} \right)\left( {1 - 2} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 2} \right)\left( {1 - 0} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 0} \right)\left( {1 - 0} \right)}} \cr & = \frac{1}{{ - 1}} + \left( {\frac{1}{{ + 1}}} \right) + \frac{1}{{ - 1 \times - 1}} + \frac{1}{{ - 1}} \cr & = - 1 + 1 + 1 - 1 \cr & = 0 \cr} $$