The Professor Cube is a 5x5x5 version of a Rubik's cube. If you already have a solid understanding of the 3x3x3 and 4x4x4 versions, then the 5x5x5 should be easy enough to solve. The cube contains the following types of pieces:
As you can see, most of the pieces are elements that have already been seen in 3x3x3 and 4x4x4 cubes. The only new type of piece is the Middle Center, 4 of each color.
Solution Outline
Solving the Centers
If you are already know some good operators for moving 4x4x4 centers, then you can solve the 5x5x5 centers by first solving the Fixed Center and Middle Centers for each face, then using the 4x4x4 operators to move the Diagonal Centers into position. Solving the Middle Centers 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 Middle Center piece. Do this at most 24 times and the Middle Centers will be complete.
The Diagonal Centers tend to take a few more moves. Initially, I used 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)
These are really handy on the 4x4x4 cube if you solve the centers last because they don't change edges or corners at all. However, for the 5x5x5 cube we are solving the centers first, so those operators are overkill.
Here's a faster operator for fixing a single Diagonal Center:
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 Diagonal Centers on R, but since we don't care about permuting the Diagonal Centers within a single face and we haven't fixed the edges yet, these side effects are harmless.
Solving the Centers without Operators
The above solution, although effective, can be a bit tedious. On a 3x3x3 cube most people solve the first layer or two intuitively before resorting to general purpose operators for the last layer. Similarly, it is possible to solve the centers without brute forcing each individual center side and center corner.
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.
Matching Up Edges
The goal here is to form edge groups by matching each middle edge with the two corresponding side edges. One very important thing to keep in mind is that face moves always move middle and side edges as a group. Thus, any number of face moves can be made without disturbing previously matched up edges. However, you still need a way to move side edges independent of middle edges, and the 4x4x4 cube's edge-3-cycle works well here:
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 perform 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). 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, and thus a different operation 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 and execute:
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.