Take any 3 digit number, abc. Writing the subtraction gives:
abc – cba = 100a -100c – a – c =99a – 99c = 99(a-c)
Since a and c differ by at least 2, the possible values of a-c are: 2,3,4,5,6,7,8
So, 99(a-c) must equal one of the following: 198, 297, 396, 495, 594, 693, 792
Now say that the result of abc – cba is def
Then the end result,P, of adding def to fed can be written as 100d + 10e + f + 100f + 10e +d
P = 100(d+f) + 20e +d + f
But from our list of possible values of 99(a-c), we can can see that the middle digit, e, is always 9. Also, the addition of the two end digits is always 9. So d+f = 9.
So our final sum P = 100(9) + 20(9) + 9
P= 900 + 180 + 9
P=1089