Problem #PRU-5070

Problem

Prove the magic trick for the number $1089={33}^2$: if you take any $3$-digit number $\overline{abc}$ with digits coming in strictly descending order and subtract from it the number obtained by reversing the digits of the original number $\overline{abc}-\overline{cba}$ you get another $3$-digit number, call it $\overline{xyz}$. Then, no matter which number you started with, the sum $\overline{xyz}+\overline{zyx}=1089$.
Recall that a number $\overline{abc}$ is divisible by $11$ if and only if $a-b+c$ also is.