If x + y + z = 6, then the value of (x - 1)³ + (y - 2)³ + (z - 3)³ is?

Solution(By Examveda Team)

$$\eqalign{ & x + y + z = 6 \cr & {\left( {x - 1} \right)^3}{\text{ + }}{\left( {y - 2} \right)^3}{\text{ + }}{\left( {z - 3} \right)^3} \cr & \therefore {\text{As, }}x + y + z = 6 \cr & {\text{Take values}} \cr & x = 1 \cr & y = 2 \cr & z = 3 \cr & \left( {1 + 2 + 3} \right) = 6 \cr & \therefore {\left( {1 - 1} \right)^3}{\text{ + }}{\left( {2 - 2} \right)^3}{\text{ + }}{\left( {3 - 3} \right)^3} \cr & = 0 \cr} $$
Now assume values in options.
Option 'D' satisfies the given relation.
Hence 'D' is correct.