If the final digit of the number is 0, then some carrying takes place when 1 is subtracted. Otherwise, no carrying happens.
If no carrying happens, call the three digits of today's number \(A\), \(B\), and \(C\). We know that \(A^3+B^3+C^3\) is one more than \(A^3 + B^3 + (C-1)^3\).
This implies that \(C^3=(C-1)^3+1\), which is only possible if \(C\) is 1.
Therefore either the final digit of today's number is 0 or the final digt of one less that today's number is 0. In both cases, we need to find a number
with the desired property whose final digit is 0: we are looking for digit \(A\) and \(B\) such that \(A^3+B^3\) is a multiple of 10.
Looking at all the cube numbers, there are a few combinations that add up to multiple of 10: