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...|..7|51.
..2|...|..9
3.9|..2|...
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.7.|.6.|..8
6..|9.5|..7
2..|.1.|.5.
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...|1..|3.5
1..|...|7..
.67|4..|...
has an interesting combination. After some normal deductions the candidate sets are:
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{8} {4} {6} {3} {9} {7} {5} {1} {2}
{7} {5} {2} {6} {48} {1} {48} {3} {9}
{3} {1} {9} {58} {458} {2} {48} {7} {6}
{45} {7} {135} {2} {6} {34} {19} {49} {8}
{6} {38} {13} {9} {38} {5} {12} {24} {7}
{2} {9} {48} {7} {1} {48} {6} {5} {3}
{49} {2} {48} {1} {7} {689} {3} {689} {5}
{1} {38} {35} {58} {258} {69} {7} {69} {4}
{59} {6} {7} {4} {235} {389} {29} {289} {1}
The four cells in rows 2/3 and columns 5/7 contain just 48, except r3c5 has candidates 458.
If we trust that the puzzle has a unique solution (we can, this is from Angus's Simple Sudoku), we can deduce r3c5 = 5 (since if not, the 48's in an X-wing cannot possibly have a unique solution).
This deduction is not needed (simpler techniques suffice). But I saw this idea mentioned in another thread, and was interested to see it actually occur.