Solution:

Here is the unique solution to the standard version described above. As mentioned earlier, the cubes can be in any order. The cube with three red faces is easiest to recognize, so start with that one. The other cubes can also be recognised by the colours which occur more than once.

There is a neat method for solving this puzzle. It shows whether there is no other solution, and works for any version of the puzzle.

1. Number the cubes. I will use the numbering shown at the top of the page.
2. You will next build a graph. Draw four points, arranged in a square which will represent the four colours.
3. Take the first cube. There are three pairs of opposite faces. For each pair, draw a line between the points of those two colours, and mark it with the number 1 to show it comes from the first cube. Note that if two opposite faces have the same colour, then you will draw a loop from the point of that colour back to itself.
4. Repeat this for the other three cubes, but now marking the lines with the appropriate numbers 2,3 and 4.
5. With the standard version this results in the following graph:
   
6. Now find two loops in this graph, such that each loop visits all four colours once, and uses lines with each number once. The two loops may not use the same lines.
7. In this example, you can find the loops:
   R - 1 - G - 4 - Y - 2 - B - 3 - R
   R - 4 - G - 3 - Y - 1 - B - 2 - R
   One loop will solve the front and back, the other the right and left sides.
8. Take cube 1, and see where it occurs in the loops. Cube 1 connects R-G and Y-B, so place the cube with Red at the front (Green at the back) and Yellow on the right (Blue on the left).
9. Do exactly the same for the other cubes, and you will then have a solution when they are stacked together. This solution is unique up to reordering the cubes or turning over the stack of cubes.

Actually, it might be possible (or necessary) to solve a set of opposite sides with several loops instead of one. In the example we needed a 4-cycle for each set of opposite sides, but two 2-cycles, or a 3-cycle with a single loop can work just as well. Together they will still have to use all 4 colours and all four cubes. The solution to Mutando is such a solution. Its graph is as follows:

As red is only connected to one other colour, it is clear that there is no loop that contains red and other colours. The solution is this:

R - 1 - R,   B - 3 - B,   G - 4 - Y - 2 - G
R - 3 - R,   B - 2 - G - 4 - Y - 1 - B.
Again this solution is essentially unique, though most cubes have two orientations that show the same colour. This puzzle is actually much easier than the standard instant insanity.

Mutando also allows you to make a 2×2×1 block, where each side has a single colour. As you can see from its graph, each cube has green opposite yellow, so the 2×2 sides must be green and yellow. Cube 3 will have red and blue on its other visible sides. Cube 1 also must show red, cube 2 has no red, so there is exactly one 2×1 red side formed by cubes 3 and 1. Cube 4 has no blue, so cube 2 must form the blue side with cube 3. The red side of cube 4 must be hidden, so the other sides can only be yellow and green. To recap:

Note that there are actually two solutions. The top can be yellow, the bottom green or vice versa.

Drive Ya Crazy has six cubes, but the same method can be used to analyse the solutions. Draw six points in a hexagon to represent the six colours, and as before connect each pair of opposite faces on each cube. This results in the following graph:

This graph is a bit more complicated than the others, but it turns out to have the nice feature that there are three sets of loops that visit each colour using all cubes:

O - 1 - R - 2 - C - 5 - B - 6 - G - 3 - Y - 4 - O
O - 2 - G - 4 - C - 6 - O,   Y - 1 - B - 3 - R - 5 - Y
O - 5 - G - 1 - C - 3 - O,   Y - 2 - B - 4 - R - 6 - Y
Any two of these sets gives a solution. Note that you may reverse the direction of one or more of the loops and this may result in a slightly different solution. An interesting feature of this puzzle is that if you have a solution (such as the one shown at the beginning of this section) then it does not matter which direction you stack the cubes in, i.e. you can put them in a row on top of, next to, or behind each other, giving a solution each way. These three solutions correspond to the three ways of choosing two of the three sets of loops above. If you choose the two pairs of short loops for the four sides then the long loop is used in the internal cube faces, and this allows the cubes to be reordered such that touching faces have the same colour.

The Dorobo five cube version also is easily analysed by the same method. It has the following graph:

The diagonals all belong to cubes 4 and 5, and the outer edged to the other three cubes. Therefore the loops we are looking for must use exactly two diagonals. There is only one loop shape that does so, which makes it easy to find the solution. Like the Drive Ya Crazy version this graph splits into three loops that visit each colour using all cubes:

G - 3 - O - 4 - C - 1 - P - 2 - Y - 5 - G
G - 2 - O - 1 - Y - 5 - C - 3 - P - 4 - G
G - 1 - O - 5 - P - 3 - Y - 4 - C - 2 - G
There are essentially three solutions, constructed as before by using any two of these loops to determine the two pairs of opposing sides. Unlike the Drive Ya Crazy, it is not possible to use the third loop to make the touching faces the same because some cubes would need to be turned over to get the matching faces together. This also means that to go from one solution to the other you must not only just stack them in a different direction but also turn around two cubes (cubes 2 and 5 to be exact).