Colorful cubes

Label the red face R, the green face G, the blue face B, the bottom face (opposite to G) X, the left back face (opposite to R) Y, and the right back face (opposite to B) Z, as pictured below.

Then X cannot be green because faces X, R, and B can be rotated to the disallowed cube, and Y cannot be red because faces Y, B, and G can be rotated to the disallowed cube, and Z cannot be blue because faces Z, R, and G can be rotated to the disallowed cube.

Assume X is blue. Then Z cannot be green because faces R, X, and Z can be rotated to the disallowed cube, and since Z cannot be blue (from above), Z must be red. If Z is red, then Y cannot be blue because faces Y, Z, and G can be rotated to the disallowed cube, and since Y cannot be red (from above), Y must be green. But if X is blue, Z is red, and Y is green, faces X, Y, and Z can be rotated to the disallowed cube. So our assumption that X is blue must be false, so X cannot be blue.

Assume X is red. Then Y cannot be green because faces X, Y, and B can be rotated to the disallowed cube, and since Y cannot be red (from above), Y must be blue. If Y is blue, then Z cannot be red because faces Y, Z, and G can be rotated to the disallowed cube, and since Z cannot be blue (from above), Z must be green. But if X is red, Y is blue, and Z is green, faces X, Y, and Z can be rotated to the disallowed cube. So our assumption that X is red must be false, so X cannot be red.

Therefore, X cannot be red, green, or blue without making a possible reorientation of the disallowed cube, so there is no way that the colors of the other three faces of the cube on the left could be painted so that it would be impossible to reorient the cube to look like the disallowed cube.

say the color opposite blue r is x, opposite red is y, opposite green is z. Going around each of the 8 corners we write the sequence (starting at an arbitrary face and going counterclockwise around it - I drew a flat cube layout to help my imagination a bit). We get 1.brg, 2.rxg 3.xyg 4.ybg 5.zrb, 6.zby 7.zyx 8.zxr. Now we must choose xyz so that none of these gets equal to the forbidden patterns bgr, grb or rbg. From 2. it follows that $x \neq b$ , from 4. that $y\neq r$ and from 5. that $z\neq g$ . Now we are left with 8 possibilities for xyz, that are listed in the first column below. Now condition 3 tells us that xy cannot be rb, because it would result in the forbidden pattern rbg.

So each possibility violates some of the requirements and thus it is not possible.