If a equals minus 3b equals 25, what is the last digit of the 1999 power of a plus the 1999 power of b?

If a equals minus 3b equals 25, what is the last digit of the 1999 power of a plus the 1999 power of b?

3^1 Mantissa is 3 3, the tail of 3^2 is 9, the tail of 3^3 is 7, the tail of 3^4 is 1, the tail of 3^5 is 3, and then it is reincarnated. Therefore, every four of the mantissa of 3^1999 is a reincarnation, so the mantissa of 3^1999 is the mantissa of 3^(1996+3), is 7, a=-3, then a^1999

3^1 Mantissa is 3 3, the end of 3^2 is 9, the end of 3^3 is 7, the end of 3^4 is 1, the end of 3^5 is 3, and then reincarnation, so it can be known that every four of the mantissa of 3 is a reincarnation, so the mantissa of 3^1999 is the mantissa of 3^(1996+3), is 7, a=-3, then a^1999

Given that the cth power of the bth power multiplication 37 of the ath power multiplication 27 of 2 is equal to 1998, where a, b, c are natural numbers, find the value of the 2006th power of (a-b-c).

1998=2*27*37
2^A*27^b*37^c =1998
2*27*37=2^A*27^b*37^c
A=1, b=1, c=1[ a, b, c are natural numbers]
(A-b-c)^2008
=(1-1-1)^2008
=1