1+x1111+y1111+z=

By expanding, we get(1+x)[(1+y)(1+z)−1]−1(z)+1(−y)=xyz(1+1x+1y+1z)Trick : Put x=1, y=2 and z=3, then211131114=2(11)−1(3)+1(1−3)=17Option (1) gives,1×2×3(1+11+12+13)=17