Can these 9 pieces be put together into a 3 x 3 x 3 cube?

Let the 27 components of the cube be {-1,0,1}^3. Then the three 7-component pieces can be arranged this way:

(1): (-1, -1, -1), (-1, 0, -1), (-1, 0, 0), (-1, 0, 1), (0, 0, -1), (1, 0, -1 ), (1, -1,-1),

(2): (1, -1, 1), (1, -1, 0), (0, -1, 0), (-1, -1, 0), (1, 0, 0), (1, 1, 0 ), (1, 1, 1),

(3): (-1, 1, 1), (0, 1, 1), (0, 0, 1), (0, -1, 1), (0, 1, 0), (0, 1, -1), (-1, 1, -1).

In other words, if we take (1) to be the original copy, then we can obtain (2) and (3) via the orthonormal transforms

((0, 1, 0), (0, 0, -1), (-1, 0, 0)), and

((0, 0, -1), (1, 0, 0), (0, -1, 0),

respectively.

I say it's not possible.

First, the 3x3 L-shaped piece can only have it's middle cube in a corner.

The 3x3 L-piece takes up 3 corners.

So three 3x3 L-pieces would take up 9 corners.

But a cube has only 8 corners.

Ah! yes, I can see that at least one 7-cube piece can have the 3x3 piece as the

middle of two faces, but I'm unable to prove a cube can or cannot be formed,

as my 3D glasses are broken.

Consider the cube, M, of the 7-er that has the most number of cubes attached to it. It's clear that that can't occupy three corner position, because each time it occupies a corner position, it takes two other corners with it, and we don't have 9 corners. Also, it's clear that one of the unused singletons will be the center cube of the large cube.

So I don't have drawing software. If you can recommend a free website where I could conveniently assemble an image, I might give that a go. Otherwise, I could withdraw my entry. I'll take it that the centers of my cubes are at integer positions in (a,b,c) where each of a,b,c are integers ranging from -1 to 1, the center cube being at (0,0,0).

Let F be the cube on the 7-er that is farthest from M. If I give the positions of M and F, that will specify the layout completely:

7-er 1: M:(-1,0,-1), F:(1,-1,-1)

7-er 2: M:(0,1,1), F:(-1,1,-1)

7-er 3: M:(1,-1,0), F:(1,1,1)

Singletons: (0,0,0), (0,-1,-1), (-1,1,0), (1,1,-1), (1,0,1), (-1,-1,1)

Remaining cubes:

7-er 1: (-1,-1,-1), (-1,0,0), (-1,0,1), (0,0,-1), (1,0,-1)

7-er 2: (0,1,0), (0,1,-1), (-1,1,1), (0,0,1), (0,-1,1)

7-er 3: (-1,-1,0), (0,-1,0), (1,0,0), (1,1,0), (1,-1,1)

Update: It seems to me that putting M at a non-corner leaves only two possible corners for the other two Ms, but neither of them work. Ie. all three Ms must be an non corners. And given that, I think the solution is forced since that means each 7-er takes out an adjacent pair of center face cubes.

Scythian, thanks for that. I'm not sure I'll have the time today to color your cube in, but I hope to get back to it.