<s>[INST] How to find expression for this tensor? First consider 1D case. Suppose I would like to compute 
\begin{alignat}{2}
 m_p = \sum_{x=1}^{N} x^p f(x), \qquad \forall\; p = 1, 2, \dots, P
\end{alignat}
I know it can be expressed as a matrix product as follows
\begin{alignat}{2}
 \underbrace{ \begin{bmatrix}
  m_1 \\
  m_2 \\
  \vdots \\
  m_P
 \end{bmatrix} }_{:=\, \boldsymbol{m}} = \underbrace{ \begin{bmatrix}
  1^1 & 2^1 & \dots & N^1 \\
  1^2 & 2^2 & \dots & N^2 \\
  \vdots & \vdots & \ddots & \vdots \\
  1^P & 2^P & \dots & N^P
 \end{bmatrix} }_{ :=\, X}  \underbrace{ \begin{bmatrix}
  f(1) \\
  f(2) \\
  \vdots \\
  f(N)
 \end{bmatrix} }_{:=\, \boldsymbol{f}}
\end{alignat}
Now generalizing this to 2D case, I know that
\begin{alignat}{2}
 m_{pq} = \sum_{x=1}^{N} \sum_{y=1}^{M} x^p y^q f(x, y), \qquad \forall\; p = 1, 2, \dots, P;\; q = 1, 2, \dots, Q
\end{alignat}
can be expressed as
\begin{alignat}{2}
 &\underbrace{ \begin{bmatrix}
  m_{11} & m_{12} & \dots & m_{1Q} \\
  m_{21} & m_{22} & \dots & m_{2Q} \\
  \vdots & \vdots & \ddots & \vdots \\
  m_{P1} & m_{P2} & \dots & m_{PQ}
 \end{bmatrix} }_{ :=\, M} =\\
 &=\underbrace{ \begin{bmatrix}
  1^1 & 2^1 & \dots & N^1 \\
  1^2 & 2^2 & \dots & N^2 \\
  \vdots & \vdots & \ddots & \vdots \\
  1^P & 2^P & \dots & N^P
 \end{bmatrix} }_{ :=\, X}
 \underbrace{ \begin{bmatrix}
  f(1,1) & f(1,2) & \dots & f(1,M) \\
  f(2,1) & f(2,2) & \dots & f(2,M) \\
  \vdots & \vdots & \ddots & \vdots \\
  f(N,1) & f(N,2) & \dots & f(N,M)
 \end{bmatrix} }_{ :=\, F}
 \underbrace{ \begin{bmatrix}
  1^1 & 2^1 & \dots & M^1 \\
  1^2 & 2^2 & \dots & M^2 \\
  \vdots & \vdots & \ddots & \vdots \\
  1^Q & 2^Q & \dots & M^Q
 \end{bmatrix}^{\top} }_{ :=\, Y^{\top}}
\end{alignat}
Now my question is how does this generalize to 3D case? That is how can I express in terms of vectors/matrices/tensors the tensor
\begin{alignat}{2}
 \begin{pmatrix}
  m_{pqr}
 \end{pmatrix}_{p=1, q=1, r=1}^{p=P, q=Q, r=R}
\end{alignat}
whose elements are given as
\begin{alignat}{2}
 m_{pqr} = \sum_{x=1}^{N} \sum_{y=1}^{M} \sum_{z=1}^{K} x^p y^q z^r f(x, y, z), \qquad \forall\; p = 1, 2, \dots, P;\; q = 1, 2, \dots, Q;\; r = 1, 2, \dots, R.
\end{alignat}
 [/INST] People usually stop writing down matrices when dealing with rank-3 tensors and higher. In your case, using the Einstein summation convention,
1D: $m_p = X_{pi} f_i$ with $X_{pi} = i^p$, $f_i = f(i)$
2D: $m_{pq} = X_{pi} Y_{qj} f_{ij}$ with $X_{pi}$ as above, $Y_{qj} = j^q$, $f_{ij} = f(i,j)$.
3D: $m_{pqr} = X_{pi} Y_{qj} Z_{rk} f_{ijk}$ with $Z_{rk} = k^r$, $f_{ijk} = f(i,j,k)$.
For the 3D case, while the $X$, $Y$, and $Z$ objects could be written as matrices, the $m_{pqr}$ and $f_{ijk}$ are not easily expressed that way. 
</s>