Or just in lines... a^n many lines of length 1 + b^n many lines of length 1 to cover the total length of c^n many lines of length 1. Or just in n-dimensional c

Or just in lines... a^n many lines of length 1 + b^n many lines of length 1 to cover the total length of c^n many lines of length 1. Or just in n-dimensional cubes from the start: 1 n-dimensional cube of side-length a + 1 n-dimensional cube of side-length b to cover the volume of 1 n-dimensional cube of side-length c. Of course, both of these don't amount to much different than just saying a^n + b^n = c^n directly. :) I don't suspect that decomposing it into a^(n - k) a^k + b^(n - k) b^k = c^(n - k) c^k for arbitrary k is particularly helpful, and, for reasons as illustrated above, I suspect the aid to intuition from thinking in terms of volume and tessellations geometrically isn't very great, but, I'm not an expert on Fermat's Last Theorem. (For all I know, something like this does get used in the proof...)