If a + b + c + d = 4, then 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)

$$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & = \frac{{1 - d + 1 - a + 1 - b + 1 - c}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - \left( {a + b + c + d} \right)}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - 4}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = 0 \cr} $$