This is a method of finding hidden unique rectangles.
Below, the  in the  is X’d out. It can be eliminated as a candidate.
If the upper left blue  cell is a , then  can’t be a candidate for this cell.
A problem arises when the blue  is any number other than . The yellow parts become candidates for .
Then the lower right blue cell would be a , since it is a . In this case, too, if the upper left, blue cell is a …
we eventually arrive at the following diagram.
But lets step back. The four blue cells create a unique rectangle with , leading to multiple solutions. Therefore, in this case, the blue cell to the upper left is a  and not a .
Hence, this  can be removed as a candidate. Otherwise, we are left with a Sudoku problem which doesn’t work.
Below is also a hidden unique rectangle. The X’d out  can be removed as a candidate from D3. (The 8s in the same row form a strong link)
If the  in the upper left B1 is an , then B3 becomes a , so the X’d out  in D3 can’t be entered there.
If the upper left B1’s  is a , we have a problem. In this case, there are only two ’s in the same row, so even if there are many candidates for D1, it will be an . Furthermore, B3 will also be an . If D3 is a , we have a unique triangle where  and  are interchangeable.
It will hence, not work as a Sudoku problem, and  can be removed as a candidate.