Fillomino with Sums: The Fillo That Smiles Back!

Question

This puzzle is a Fillomino, with the special variant rule of Sums ("Cages") being applied. You might also notice that this particular fillomino is giving you a smile! This is my first time making a Fillomino, especially one with variant rules, so I hope you enjoy.

Standard Fillomino rules: (paraphrased from Melon's Puzzle Pack Volume IV)

In Fillomino, the goal is to divide all of the grid squares into polyominoes, subject to the following constraints:

1. Every number in the grid must be contained in a polyomino containing that quantity of squares.
2. No two polyominoes containing the same quantity of squares may share an edge.
3. A polyomino may contain one, more than one, or none of the numbers originally given.

Sum rule variant: (paraphrased from Melon's Puzzle Pack Volume IV)

Sum: In addition to the usual rules, the grid contains some cages. The number at the top left of each cage gives the sum of all numbers that appear inside of it. Numbers may be repeated in cages.

Now, here is the puzzle:

And here is a Penpa link to solve it in your browser, if you'd like. (Quick Penpa tip if you haven't used it before: Placing numbers uses "Mode -> Number", and placing walls uses "Mode -> Edge.")

Answer 1

First, the basic deductions from extending regions:

Next, look at some sum cages.

The very lower right cell must not be alone, so it must be connected to the cell above it. That makes that number 3.

Similarly, the very lower left cannot be a 3, because it would run into the 3s above it. It also can't be a 5, because then the other cell in the pair would be a 1. That means the lower left cell must be a 1.

Next, the upper left:

The 9 group in column 3 doesn't have enough room in the upper left. So it must join up with the 9 group on top, through the cell R2C4.

This lets us resolve the upper left area entirely.

...And from here, I believe there are multiple solutions.

Unless I've made a mistake (which very well could be the case), this demonstrates at least two solutions to the puzzle. (I've highlighted 1s, 2s, and 3s for easier checking.)