X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit's place of the numbers $$(408X)^{63}$$ and $$(789Y)^{85}$$ are the same. What will be the possible value(s) of (X + Y)?

All numbers from 1 to 9 repeat their last digits over a cycle of 4. 
63 can be written as 4k+3.
85 can be written as 4k+1. 
Some number's third power should yield the first digit as some number's first power.
$$2^3$$ will yield $$8$$ as the last digit (2 and 8 is a possible solution).X+Y = 10
$$3^3$$ will yield $$7$$ as the last digit (3 and 7 is a possible solution).X + Y = 10
$$4^3$$ will yield $$4$$ as the last digit and hence, can be eliminated. 
$$5$$ and $$6$$ yield $$5$$ and $$6$$ respectively as the last digit for any power and hence, can be eliminated.
$$7^3$$ will yield $$3$$ as the last digit (7 and 3 is a possible solution). X+Y=10.
$$8^3$$ will yield $$2$$ as the last digit. 8+2 =10.
As we can see, X+Y = 10 in all the cases. Therefore, option B is the right answer.