How do you express the product of three matrices as a weighted sum of outer products? How do you express the matrix ABC as the weighted sum of outer products of vectors extracted from A and C and with weights coming from matrix B?
I'm not sure how to approach this problem. I tried the outer products of the columns of A with the rows of C, with the entries $B_{ij}$ of the middle matrix serving as the weights. Since B has as many columns as C has rows and as many rows as A has columns, we would have as many outer products in our sum as B has entries. But it didn't work out.
 A: Let us consider the $2 \times 2$ case for the sake of simplicity:
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}t&u\\v&w\end{pmatrix}\begin{pmatrix}e&f\\g&h\end{pmatrix}=t\begin{pmatrix}a\\c\end{pmatrix}\begin{pmatrix}e&f\end{pmatrix}+u\begin{pmatrix}a\\c\end{pmatrix}\begin{pmatrix}g&h\end{pmatrix}+...$$
Indeed, if you denote by $C_1,C_2$ the columns of the first matrix and by $R_1,R_2$ the rows of the 3rd matrix, you just do block products (all with compatible dimensions) like this:
$$\begin{pmatrix}C_1&C_2\end{pmatrix}\begin{pmatrix}t&u\\v&w\end{pmatrix}\begin{pmatrix}R_1\\R_2 \end{pmatrix}=\begin{pmatrix}(C_1t+C_2v)&(C_1u+C_2w)\end{pmatrix}\begin{pmatrix}R_1\\R_2\end{pmatrix}$$
$$=(C_1t+C_2v)R_1+(C_1u+C_2w)R_2=tC_1R_1+vC_2R_1+uC_1R_2+wC_2R_2$$
This process can be extended to any dimensions, not only to $n \times n$ square matrices but more generally to products of rectangular matrices as long as their dimensions are compatible, i.e., with dimensions:
$$(m \times n) \times (n \times p) \times (p \times q)$$
A similar question asked in the other direction.
