This page summarizes some of the techniques I use for solving the 5x5x5 cube. It is not a step-by-step walkthough of a solution.

Pieces are described by the faces they are closest to. An upper case letter means the piece is actually on the face, while a lower case letter indicates that it is one space in from the face. When a piece is in the middle, no letter is used for that particular dimension. So, UFR is the top-right corner on the front face. Moving to the left, we have UFr, UF, UFl, and UFL.

- 8 Corners (example UFR): these move just like the corners on all
Rubik's Cubes

- 12 Middle Edges (example UF) : these move like the edges on a 3x3x3 cube
- 24 Side Edges (example UFr): these move like the edges on a 4x4x4 cube
- 24 Center Corners (example uFr) these move like the centers of a
4x4x4 cube

- 24 Center Sides (example uF): unique to the 5x5x5 cube

- 6 Centers: these never move in relation to one another (like the
3x3x3 cube centers)

- Solve the 9 center pieces on each face (see below)

- Match Side Edge pieces with their corresponding Middle Edges (see below)
- Solve the cube as a 3x3x3 cube using only face moves (no inner slice moves)

Solving the Center Sides without preserving any other pieces is actually quite simple. The following two sequences should be enough:

r' F r : moves uF to Ur

r2 D r2 : moves Df to Ur

Remember that faces can be turned without disturbing any center pieces and you can also turn the entire cube to face any way you want, so it is quite easy to arrange things such that one of the above two sequences solves a Center Side piece. Do this at most 24 times and the Center Sides will be complete.

The Center Corners tend to take a few more moves. Initially, I used the following operators such as:

- double-center-swap: U2 b'
u'd b U2 b' u d' b (use to exchange Ubr/Ubl with Fur/Fdr)

- single-center-swap: U2 b' u' b U2 b' u b (use to exchange Ubl with Fur)
- double-center-swap2: r2 u2 r2 u2 (exchange Ufr/Ubr with Dfr/Dbr)

Here's a faster operator for fixing a single center corner.

center-corner-move: d R d' R d R2 d' - move Rdf to Fdr

Note that it has some side effects. It cycles some side edges on R and cycles side corners on R, but since we don't care about permuting the side corners within a single face and we haven't fixed the edges yet, these side effects are harmless.

It is quite east to work with 2 faces at a time without distrubing the centers on the other 4 faces. The key is to do mostly face moves and always match a slice move with its inverse. Between the slice and its inverse you have to be careful to only turn a face that consists entirely of pieces that you are willing to move. For example, when working on the U and F centers you could do something like this:

U2 F r' F2 r U

Actually, the stuff before r' and after r is irrelevant - it can be any combination of U and F moves that you want. Since we're only concerned with the centers, the order of the moves doesn't even matter since upper centers are only affected by U and front centers are only affected by F. The important part is that between r and r', you restrict yourself to turning F.

edge-3-cycle: r (U' R U) r' (U' R' U)

This will cycle DFr, UFr, FRu. Actually, you can skip the last three moves and just do r U' R U r', because the last three moves are face moves which have no effect on matching of edges. Their only purpose is to restore edge groups and corners to their original position. Since we only care about matching up the edges, and not where the edge group (or corners) wind up, we can skip those last moves.

I usually just pick a piece at DFr, and use face moves to move the corresponding middle edge to UF. Make sure you get UF flipped the right way (the front face of DFr should match the up face of UF). Then examine UFr and put the corresponding middle edge at FR (the F face of UFr should match the R face of FR). Then execute the edge-3-cycle and you are 2 pieces closer to matching all of the edge groups.

It is possible to wind up with only two unmatched side edges that need to be swapped with one another. No amount of manipulation with edge-3-cycles will be able to solve this problem. Another operatoin is required. Basically, you need to create an odd permutation of the edges. Once you do this, you should then be able to perform one or more 3 cycles to clean things up. My preferred way of solving this problem is to put the two problematic edges at uFR and uBR, then do this:

edge-4-cycle: R2 u R2 u R2 u R2 u R2 u R2.

This cycles uFR, uBR, uBL, and uFL. Since uFR and uBR needed to be swapped, after the 4-cycle, exacly three side edges will be out of place and an edge-3-cycle should be able to fix them.