The method I show here is an alternative to a simpler and easier method known as the peddle-flip method.
In the steps that follow, we are going to let A=LFU, B=RFU, C=RBU and D=LBU.
Solve the edges. Start with one side of the cube, then do the other. A simple tri-cycle of the edges finishes the final layer.
From here on we only use sequences of the form X,Y,Xi,Yi to tri-cycle slivers about the cube while preserving the solved edges. Solving the puzzle from here on means mastering these tri-cycles in all variants, forward and backards.
In this step we temporarily position the centers. This is just to see if we have parity. If there becomes two remaining centers that need to be swapped, use X,2|R|,X with X=2[C,Di],C. Then, though not necessary, you can temporarily solve the center positions again to see that they solve without a problem.
I now like to solve all the spike-pairs. There are 2 sets of spike-pairs, each having 6 pairs, 1 per face. There are 2 spike-pairs per face, one in each set. Make a setup move to complete a pair (by matching 2 spikes of the same color), then tri-cycle an unsolved pair into the position of the solved pair, therey-by kicking out the solved pair into a safe location, then undo the setup move. This strategy works for a single set of spike-pairs up until the last 2 or 3 remaining pairs. If there are 2 remaining pairs, use a similar strategy to turn it into 3 remaining unsolved pairs. If there are 3 remaining pairs, solve them all simultaneously. After positioning the remaining pairs strategically, make a setup move, then a tri-cycle, just as you've been doing all along, so that undoing the setup-move completes the step for a single spike-pair set. It's really the same basic sequence strategy in all cases. Solve both spike-pair sets.
Easily solve 3 of the 6 sides using sliver tri-cycles. The next step will address the reamining 3 sides. I always make the 3 remaining sides a strip (i.e., not pair-wise adjacent.)
The key move here is Bi,2|R|,Ci,Ai,C,A,2|R|,2B,|Ri|,B,D,Bi,Di,|R|,2Bi,2|R|,C,A,Ci,Ai,2|R|,B,|Ri|,B,D,Bi,Di and a varient of this for the other spike-pair set. That's a long sequence, but it's not hard to memorize the basic idea. What we're essentially doing is turning 3 solved spike-pairs into 3 unsolved spike-pairs, then resolving them in such a way that when we try positioning them again in the final 3 faces, they succeed in going into their proper places next to the edges.
Once all the spikes are positioned properly, all that may remain is to tri-cycle 3 centers. The center tri-cycle is given by 2[C,A,Ci,Ai,(D,B,Di,Bi),C,Ai,Ci,A],D,B,Di,Bi.