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As far as I can tell, if you can solve the order 3 and order 4 cube,

you should be able to solve the order 5 with no additional fiddling, even if

you only know cookbook solutions.

Spoiler follows:

I solve the off-center edges first (just like in the order 4 cube--

the transformations are identical), then the corners (exactly like all other

orders, from 2 through 4), then the center edges (exactly like the order 3

cube, just treat the two edge faces as attached and you have an order 3

cube). All that's left are the eight centers. Four of these can be solved

exactly as in the order 4, and if you can't generalize your cookbook solution

to solve the remaining 4 you have no business cubing.

I suspect this is why there are (and will probably never be) cubes of

orders greater than 5. I believe (though have not proved) that the 5 cube

contains all the complexity that is possible. Adding more cubies would only

increase the amount of time needed to solve.

On the other hand, I would be willing to pay a fair amount of money for

an order 21 cube. :-)

Ronnie