I'm currently working with resultants, which I define as follows: let $k$ be a field, $V$ be a 2-dimensional vector space over $k$, and let $S^dV^*$ be the $d$-th symmetric power of dual space, i.e. linear span of all homogeneous polynomials of degree $d$ in two variables $x_0, x_1$ over $k$. Define for $f \in S^mV^{*}$, $g \in S^nV^*$ a map $$\theta_{f,g} : S^{n-1}V^* \times S^{m-1}V^* \longrightarrow S^{m+n-1}V^*$$ as $\theta_{f,g}(a,b) = af + bg$. Then matrix of $\theta_{f,g}$ in standard monomial basis is precisely the Sylvester matrix of $f$ and $g$, and thus I define resultant of $f$, $g$ as $$R_{f,g} = \det\theta_{f,g}.$$ It clearly agrees with all other standard definitions of resultants I've seen.
Proposition.
Now, I've been able to show an interesting identity regarding resultants (up to signs): $$R_{f, gh} \cdot R_{g, h} = R_{g, fh} \cdot R_{f, h} = R_{h, fg} \cdot R_{f, g}$$ by considering the following map, which I call $\theta_{f, g, h}$: \begin{align} \theta_{f,g,h}: S^{m-1}V^* \times S^{n-1}V^* \times S^{l-1}V^* &\longrightarrow S^{m+n+l-1}V^*, \\ (a, b, c) &\longmapsto agh + bfh + cfg. \end{align}
Proof:
The aforementioned identity holds as $\theta_{f,g,h}$ can be decomposed into composition of $\mathrm{id} \times \theta_{g, h}$ and $\theta_{gh, f}$: $$(a, b, c) \overset{\mathrm{id} \times \theta_{g, h}}{\longmapsto} (a, bh + cg) \overset{\theta_{gh, f}}{\mapsto} agh + (bh+cg)f = \theta_{f, g, h}(a, b, c).$$ Thus by taking $\det$ I obtain that $$\det \theta_{f, g, h} = R_{gh, f} \cdot R_{g, h},$$ and, by considering analogous decompositions, we obtain the desired identity.
Question. Now I really want to call $\det\theta_{f,g,h}$ a resultant of three bivariate homogeneous polynomials. Yet I have never seen such thing defined anywhere. If this is true, then, I believe, the following also must be true: $$\det\theta_{f,g,h} = R_{f,g} \cdot R_{g,h} \cdot R_{f,h}.$$
However, I got stuck trying to prove this. Is this indeed true? I would really appreciate any help regarding this problem. Thank you!
P.S. The question arose while studying intersection theory of plane curves, where one of the possible ways to define intersection number of two curves $F$ and $G$ at a point $w$ is to take resultant of $F\vert_{(pq)}$, $G\vert_{(pq)}$ which stand for restrictions of our curves onto the varying line $(pq)$, and consider it as a polynomial in $p$ and $q$. Then the order at which it vanishes for a generic line passing through $w$, is our intersection number.
Now, for intersection numbers we have $(F,GH)_p = (F, G)_p + (F, H)_p$, and this equality easily follows from $R_{f,gh} = R_{f,g}\cdot R_{f,h}$. I was looking for an elegant way to prove this equality and, whilst doing so, came up with all of the above. If the hypothesis my question is about is true, then it indeed solves my problem.