It's known at this point that the truncated cuboctahedron (i.e. great rhombicuboctahedron) and truncated icosidodecahedron (great rhombicosidodecahedron) can't actually be formed by truncating the cuboctahedron and icosidodecahedron. Some faces would end up irregular.
It's troubling that some try to solve this naming issue by calling them canti-/omnitruncated cube/octahedron/dodecahedron/icosahedron. And that stems from the accepted transformation being called canti- or omnitruncation, which is officially defined as the truncation of the rectification - which is again, very very wrong, for the above mentioned reasons.
But so far all my searches for how to construct these Archimedean solids continue to tell the truncation myth - a select few will say to fudge or "adjust" or "distort" things to get the result you want, but that's hardly mathematical.
One source https://www.mi.sanu.ac.rs/vismath/zefirosept2011/_truncation_Archimedean_polyhedra.htm does explain mathematically how to derive polyhedra that will form our desired results by truncation. It also shows that rectification of the cuboctahedron and icosahedron don't actually form the (small) rhombicuboctahedron & rhombicosidodecahedron, and therefore 2 rectifications do not equal a cantellation or expansion, and derives starting polyhedra for those as well.
But it doesn't explain why. Do the derived ratios have any significance? Are the starting non-Archimedean solids that get truncated & rectified to form our 4 desired Archimedean solids special for any reason other than this?