I can't seem to comment on this question Packing problem cube and cuboids but it is related. I just want to know what is the specific method used in the answer so I can try to replicate it for my own question. Specifically, how do you get BRICKS(i,i,i)=0 and BOX(i,i,i)=−2+2i using the values of 10x10x10 and bricks of 1x1x4. How do you get $x^ay^bz^c(1+x+x^2+x^3)$ ? Is this a specific formula?
The $i$ is the square root of $-1$, so the sums come out based on that.
Specifically note that $1 + i + i^2 + i^3 = 0$.
Using $i$ makes calculating a specific value of the polynomials nice because a lot of stuff cancels. In this case the counterexample is clear, and leads to the answer.
What I gather for the polynomial expression for the cube is that each space in the cube is "indexed" by the existence of a term with the appropriate powers on $x,y,z$. So the polynomial for the whole cube has $1,000$ terms from $1$ to $x^9y^9z^9$.
As for $x^ay^bz^c(1 + x + x^2 + x^3)$ that's the representation of a $1 \times 1 \times 4$ brick aligned along the $x$ axis. It starts at $(a,b,c)$ and takes up four squares in that direction. So such a brick starting at $(2,3,4)$ would have the polynomial representation $x^2y^3z^4(1 + x + x^2 + x^3)$.
If you had, say, a $1 \times 2 \times 3$ brick, one orientation would be
$$x^ay^bz^c(1 + x + x^2)(1+y).$$
There would be five others.
Or a $2 \times 2 \times 2$ brick is
$$x^ay^bz^c(1 + x)(1+y)(1+z),$$
and that's the only one you need.