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Decryption key is the first row (string of 8 digits).

100?(4) = 8/9(4) must include [0] (otherwise min sum = 1+2+3+4 = 10 > 9)
If you see two 100?(4) cages occupying the same two rows/columns then you can safely eliminate [0] from any other cell in these two rows/columns.

Notation:
J1 = Jigsaw at R1C1, J2 = Jigsaw at R1C4, J3 = Jigsaw at R1C6, J4 = Jigsaw at R3C3
J5 = Jigsaw at R4C7, J6 = Jigsaw at R5C3, J7 = Jigsaw at R6C1, J8 = Jigsaw at R6C7

1:
R3C1 0?b(2) = 1(2) = {01} (C1,R3C34)
LOL R123: R12C45+R3C3456 = 28(8) = {01234567}
R2C4 1011?b(4) = 22(4) = {4567} (R12C45+R3C3456)
R3C2 10?b(2) = 4/5(2) = [13/23/32] ([3] R3)
--> R3C2 = [1/2/3]
--> R3C1236 = {0123} (R3)

2:
R5C1 110?b(2) = 12/13(2) = {57/67} ([7] C1)
R1C1 110?b(3) = 12/13(3)
--> R1C2 <> [0] (or max sum = 0+5+6 = 11 < 12)
(Also R1C2 <> [1] with C1 12/13(2) blocking R12C1 = {56}, but I will ignore this for now.)
R1C3 11?b(2) = 6/7(2)
--> R1C3 <> [0/1/2/3] (or max sum = 2+3 = 5 < 6)
R2C2 100?b(2) = 8/9(2) <> [0] (or max sum = 0+7 = 7 < 8)
Hidden single J1: [01]

3:
[0] of R12C45+R3C3456 locked in R1C45 (R1,J2)
LOL C678: R56C5 match R34C6, <> [0]
R4C4 100?b(4) = 8/9(4) must include [0] (or min sum = 1+2+3+4 = 10 > 9)
--> [0] locked in R45C4 (C4,R5C2)
Hidden single R1: R1C5 = [0]

4:
LOL R678: R6C3456+R78C45 = 28(8) = {01234567}
R6C5 1?b(2) = 3(2) = {12} (C5,R6C3456+R78C45)
LOL C678: R34C6 match R56C5, <> [12]
--> R4C6 <> [2]
LOL R123: R3C34 match R4C56, <> [2]
--> R3C3 = [3]

5:
LOL C123: R3456C3+R45C12 = 28(8) = {01234567}
--> R45C23 <> [1/3]
R4C2 11?b(2) = 6/7(2)
R5C2 100?b(2) = 8/9(2) <> [0] (or max sum = 0+7 = 7 < 8)
--> Total sum of these 2 cages = 14/15/16
Innies R3456C3+R45C12: R5C1+R6C3 = 8/9/10
--> R5C1 <> [7] (R6C3 <> [1/2/3])
Step 2: R6C1 = [7]
--> R5C1+R6C3 = 9/10 = [54/64]
--> R6C3 = [4]
Hidden single R6C3456+R78C45: R6C6 = [0]

6:
LOL R678: R5C34 match R6C56, both = [20]
LOL C123: R34C4 match R56C3, = [42]
LOL R123: R4C56 match R3C34, = {34} (R4,J2)
J2: R1C4+R3C6 = [12]
R3C2 10?b(2) = 4(2) = [13]
R6C5 1?b(2) = 3(2) = [21]
R6C3 10?b(2) = 4(2) = [40]
R8C3 10?b(2) = 4(2) = [13]

7:
R6C2 101?b(2) = 10/11(2) = {46/56} ([6] C2,J7)
R5C2 100?b(2) = 9(2) = [72]
R4C2 11?b(2) = 6(2) = [06]
R4C6 100?b(2) = 8/9(2) = [35/45]
--> R4C7 = [5]

Mop-up:
R2C6 100?b(4) = 9(4) = [1026]
R3C8 110?b(2) = 12(2) = [57]
R1C3 11?b(2) = 6(2) = [51]
R2C2 100?b(2) = 9(2) = [27]
R7C1 100?b(3) = 9(3) = [324]
R7C8 110?b(3) = 13(3) = [670]
R6C2 101?b(2) = 11(2) = [65]
R6C4 110?b(2) = 12(2) = [57]
R6C6 100?b(4) = 9(4) = [0342]
R1C9 100?b(3) = 9(3) = [423]
R1C1 110?b(3) = 13(3) = [634]
R1C5 11?b(2) = 7(2) = [07]
R2C4 1011?b(4) = 22(4) = [6547]
R4C6 100?b(2) = 8(2) = [35]
R4C4 100?b(4) = 9(4) = [2403]
R5C1 110?b(2) = 12(2) = [57]
R5C6 11?b(2) = 7(2) = [61]
R5C8 10?b(2) = 5(2) = [41]
R8C5 101?b(2) = 11(2) = [65]

I was fixated on the outies of J7 when trying to eliminate 0 from r6c3. I didn't think to consider the innies from the hidden octet in c123.