Now a matrix has two sides, [?]. But when the array is extended to a cube, there will be 6 sides.
If a number has \(3\) prime factors \(\{A,B,C\}\), then there are the six selections, as permutations of a pair from this set. Thus, where the simple rules of left and right operations worked before, now there will be more rules. Let us draw up a notation to deal with this formally.
For a number \(N\), which is an infinite cubic array, \(N_{ijk}\), whose construction is \(p_1\otimes p_2 \otimes p_3\). We will have, for matrices \(M_{ij}\) which represent constructions \(q_1 \otimes q_2\), the following operations \[[A,B]\odot^1[A,B,C]=[C] \\ [A,C]\odot^2[A,B,C]=[B] \\ [B,C]\odot^3[A,B,C]=[A] \\ [B,A]\odot^4[A,B,C]=[C] \\ [C,A]\odot^5[A,B,C]=[B] \\ [C,B]\odot^6[A,B,C]=[A]\]
As a test, there can be \(2p\otimes 3p\otimes 5p\), and the matrix \(2p \otimes 3p\), Then by summing over two indicies, there will be left the vector representing \(5p\).
The only element that is one in the cube will be element \(N_{123}\). The only element that is one in the matrix will be \(M_{12}\). Thus we must have the product in the form
\[\odot^1 := \sum_{i}\sum_{j} M_{ij}N_{ijk}=v_{k}\]
and as a result, \(v=[0,0,1]\).
For the matrix \(2p \otimes 5p\), the only one in the matrix lies in element \(M_{13}\). Thus
\[\odot^2 := \sum_{i}\sum_{k} M_{ik}N_{ijk}=v_{j}\]
and as a result, \(v=[0,1,0]\).
For the matrix \(3p \otimes 5p\), the only one in the matrix lies in element \(M_{23}\). Thus
\[\odot^3 := \sum_{j}\sum_{k} M_{jk}N_{ijk}=v_{i}\]
and as a result, \(v=[1,0,0]\).
For all others we take the transpose of the matrix
\[\odot^4 := \sum_{i}\sum_{j} M_{ji}N_{ijk}=v_{k} \\ \odot^5 := \sum_{i}\sum_{k} M_{ki}N_{ijk}=v_{j} \\ \odot^6 := \sum_{j}\sum_{k} M_{kj}N_{ijk}=v_{i}\]