This is how I have tried: Initial stage: One triplet of the form $(n,n,n)$.
Second stage: Decompose original triplet into two triplets by splitting one of the elements of $(x,y,z)$ into two parts at a time according to the following rule: $(x,y,z)$ is replaced by $(x,y,z-r)(x,y,r)$, where $0 < r < z/2$. Each resulting triplet must be distinct in element composition when comparing with the rest of the triplets in the set. Sets of the same element composition including the case $r = z/2$ are named duplicates and set aside to reconsider when forming the next term.
Third stage: Apply the same rule to one of the triplets of a set of triplets of the second term at a time to create another two distinct triplets. At this stage consider the duplicates of the second term and apply the same rule to decompose one of the identical triplets into two triplets of distinct element composition. Repeat the process until further decomposition is not possible.
Finally sets of a decomposition of same size (equal number of triplets) must be unique with distinct composition of triplets and duplicates must be deleted.
I believe this algorithm generates only the geometrically feasible triplets in other words only the valid sets of cuboids. But when it comes to decomposing into five parts there exists sets which cannot be generated by the supposed algorithm. One such example is the set, $(5,4,3)(5,4,1)(5,3,1)(5,2,1)(5,2,2)$
I have tried to include such cases by applying the algorithm backwards to combine two triplets and again forward to decompose into two parts but it gives sets which are not geometrically feasible. Therefore we need an exclusion stage to remove invalid cases.
I need your kind assistance to find out whether there is any way to remove those invalid cases.
Do we have any flaws when applying this approach in partitioning cube into three or four parts?
I would like to share the outcomes which are published on Online Encyclopedia of Integer Sequences related to this problem. Relevant serial numbers are A381847, A384311, A384479 and A386296.
