Resultants for compactly represented product form polynomials? Typically computing resultnt of $n+1$ different $n+1$-variate homogeneous polynomials takes $O(poly(\prod_{i=1}^{d_{i}}))$ time where $d_i$ is degree of $i$th polynomial. In certain cases the polynomial can be compactly represented compact product forms.
Given $2n$ homogeneous polynomials in $2n$ variables
$$f_j(x_1,\dots,x_n,y_j)=\prod_{i\in\mathcal T_j}(x_i-y_j)$$
$$g_i(x_i,y_1,\dots,y_n)=\prod_{j\in\mathcal U_i}(x_i-y_j)$$ where $\mathcal T_j,\mathcal U_i$ are subsets of $\{1,\dots,n\}$ expressing the resultant for all the $f_j$'s and the resultant for all the $f_j,g_i$'s as determinant of Macaulay matrices directly takes exponential time for both. Since the polynomials are structured is there a closed form for the two resultants or at least can the latter resultant of both $f_j,g_i$ be computable by a determinant of polynomial size?
 A: Too long for a comment so I am posting this as an answer.
It would help if you said precisely what notion of resultant you are using and which variables are being eliminated. There are at least two, off course closely related, notions of resultants.
If $f_1,\ldots,f_{n+1}$ are $n+1$ non-necessarily homogeneous polynomials in $n$ variables $x_1,\ldots,x_n$, then one can form the nonhomogenous resultant ${\rm res}(f_1,\ldots,f_{n+1})$ which is a polynomial in the coefficients of the $f$'s and where the variables $x_1,\ldots,x_n$ have been eliminated.
If $F_1,\ldots,F_{n+1}$ are $n+1$ homogeneous polynomials in $n+1$ variables $x_1,\ldots,x_{n+1}$, then one can form the homogeneous resultant ${\rm Res}(F_1,\ldots,F_{n+1})$ which is a polynomial in the coefficients of the forms $F$ and where $x_1,\ldots,x_{n+1}$ have been eliminated.
I could not figure out from the question which notion you are using.
In any case, the homogeneous resultant satisfies tons of identities, e.g., the multiplicative property
$$
{\rm Res}(G_1 H_1,F_2,\ldots,F_{n+1})=
{\rm Res}(G_1,F_2,\ldots,F_{n+1})\ \times\ 
{\rm Res}(H_1,F_2,\ldots,F_{n+1})\ .
$$
The same is true if any other entry of the resultant factorizes as a product of two lower degree forms.
By iterating the identity, one can see that if each $F$ is a product of linear forms, the resultant ${\rm Res}(F_1,\ldots,F_{n+1})$ is the product of $d_1\cdots d_{n+1}$ resultants of linear forms where $d_i={\rm deg}(F_i)$. The latter are just the ordinary determinants of the matrices made by the selected linear forms, which should be easy to compute.

Edit post @Freeman.'s explanation:
OK, it seems the OP wants the homogeneous resultant
$$
R={\rm Res}(f_1,\ldots, f_n,g_1,\ldots,g_n)
$$
obtained by eliminating all the variables. 
Let $[n]$ denote $\{1,\ldots,n\}$. Let $A$ be the set of all functions $\alpha:[n]\rightarrow [n]$ such that $\alpha(j)\in\mathcal{T}_j$ for all $j\in [n]$. Likewise, let $B$ be the set of all functions $\beta:[n]\rightarrow [n]$ such that $\beta(i)\in\mathcal{U}_i$ for all $i\in [n]$.
Then, by my previous remarks, $R$ is a product of determinants of $2n\times 2n$ matrices which can be described in $2\times 2$ block form as follows.
$$
R=\prod_{(\alpha,\beta)\in A\times B}{\rm det}\left(\begin{array}{cc}
M_{\alpha} & -I_n\\
I_n & -M_{\beta}
\end{array}\right)\ .
$$
For any map $\phi:[n]\rightarrow [n]$, I used the notation $M_{\phi}$
for the $n\times n$ matrix which has, in each row $i$, a $1$ at position $\phi(i)$ and zeros otherwise. Of course, $I_n$ is the identity matrix.
Note that
$$
{\rm det}\left(\begin{array}{cc}
M_{\alpha} & -I_n\\
I_n & -M_{\beta}
\end{array}\right)={\rm det}(I_n-M_{\alpha}M_{\beta})=
{\rm det}(I_n-M_{\beta\circ\alpha})\ .
$$
From this it follows that if there exists $\alpha,\beta$ such that $\beta\circ\alpha$ has a fixed point in $[n]$, then $R=0$.
I would have to think more about the combinatorics of the sets $\mathcal{T}$ and $\mathcal{U}$ but I suspect that $R$ would simply vanish most (all?) of the time.

Another edit: In fact the constant vector is in the kernel of these matrices so all of the above determinants vanish. So $R$ is always zero (except if one of the $\mathcal{T},\mathcal{U}$ set is empty). In retrospect, this was obvious from the beginning because the projective system of equations has a nontrivial solution: take all the $x,y$ to be equal to the same nonzero number.
