I just looked it up and it looks quite difficult. It would be an interesting challenge but I don't know how you would solve it. Sorry, can't help you much.
I would assume you could just use basic 2x2x2 algorithms, but accounting for the conjoining cube as one corner. The middle two may take a bit more trial and error.
This really seems more of a challenge in the task of trying not to break it.
At first it looked like it may cause confusing collisions on the two inner cubes, but on further thought, it seems that it would be just like solving for 4 2x2x2 cubes.