Cage cloning is when two cages must have the same combination. This is similar to cell cloning (a thread on this starting soon) where two cells must have the same candidates (and of course eventually, the same placement). Cage cloning is much less common but even more stunning!
The first time I saw this done was in Mike's (mhparker) walkthrough for A49v2 here. See how productive step 38 is! Step 35 also relies on cage cloning but is not important to the solution. Note, Mike's WT also uses cell cloning (step 37) to make the crucial elimination of no 4 in r46c8. I've included that step as well for your enjoyment.
30. Important observation: 9(2) at R8C9 must contain 1 of {1234}
30a. Therefore, whichever combination it contains, it directly determines the combinations
{14|23} in the two 5(2) cages at R1C9 and R7C7
30b. These two 5(2) cages are therefore synchronized (i.e., must contain the same combination)
30c. Thus, they also lock the same 2 digits into R456C8 (i.e., 2 of these 3 cells must sum to 5)
(no eliminations yet)
35. 2 in C9 already locked in 5(2) at R1C9 and 9(2) at R8C9
35a. Due to synchronization of 5(2) cages (step 30b), 2 must therefore also be locked in N9 in
5(2) at R7C7 and 9(2) at R8C9 -> no 2 elsewhere in N9
37. R46C8 = {2..} (step 33), i.e., one of these 2 cells is a 2
37a. Innie/outie difference(N6) (R46C8 - R5C6 = 2) -> other cell in R46C8 = R5C6
37b. Thus, R46C8 cannot contain any candidate (apart from 2) not in R5C6
-> no 4 in R46C8
38. Hidden 5(2) pair in R456C8 (step 30c) must now be {23} (4 unavailable)
38a. Therefore (steps 30b, 30c) 5(2) at R7C7 = [23] and 5(2) at R1C9 = {23}, locked for C9 and N3
Many thanks to Andrew for good suggestions to improve this post.
Cheers
Ed