Riddles, brain puzzles and mathematical problems

[BAndrew]

We have a 3-digit number, that we call abc.

If we shuffle the numbers around, and get the combination acb, the number grows by 9.
If we instead shuffle them around and get the combination bac, the number grows by 90.

How much larger/smaller would the combination cba be, in relation to the original number, abc?

Mathematical proof is needed!

Spoiler below!

acb = abc + 9:

100a + 10c + b = 100a + 10 b + c + 9 <=>
10c + b = 10b + c + 9 <=>
9c = 9b + 9 <=>
c = b + 1 <=>
b = c - 1 (1)


bac = abc + 90:
100b + 10 a + c = 100a + 10 b + c + 90 <=>
90b = 90a + 90 <=>
b = a + 1 (2)

(1),(2) => c-1 = a+1 <=>
c-a = 2(3)



cba = 100c + 10b + a = 100a +10b + c - 99a + 99c =
100a + 10b + c + 99(c-a) = with (3)
= 100a + 10b + c + 198

cba = abc + 198

(I hope I haven't made any stupid mistake here)