Here is something which came up in my algebraic geometry class, and I'm wondering if it has a deeper explanation. Let $F(w,x,y,z) = w^3+x^3+y^3+z^3$ and let $X$ be the cubic surface in $\mathbb{P}^3$ where $F$ vanishes. As is well known, there are $27$ lines on $X$ (in characteristic $\neq 3$).


I had my students verify that the obvious equations on the Grassmannian $G(2,4)$ define the $27$ points corresponding to these lines as a reduced scheme. This is easiest to do in an affine cover. For example, consider the affine chart of lines of the form 
$$\mathrm{RowSpan} \begin{pmatrix} 1 & 0 & p & q \\ 0 & 1 & r & s \end{pmatrix}.$$
This chart contains $18$ of the $27$ points in question. The obvious equations come from setting equal to $0$ the coefficients of $t^3$, $t^2 u$, $t u^2$ and $u^3$ in
$$F(t,u,pt+ru,qt+su).$$
I.e.
$$1 + r^3 + s^3 = 3 p r^2 + 3 q s^2= 3 p^2 r + 3 q^2 s= 1 + p^3 + q^3=0. \quad (\ast)$$


To verify that they define a radical ideal, one must check the Jacobian condition: I.e., that
$$\det \begin{pmatrix}
0 & 0 & 3 r^2 & 3 s^2\\
3 r^2 & 3 s^2 & 6pr & 6 qs \\
6pr & 6 qs & 3 p^2 & 3 q^2 \\
3 p^2 & 3 q^2 & 0 & 0 \\
\end{pmatrix} \neq 0$$
at each of the $18$ roots of $(\ast)$. (Actually, I only assigned them to do one root; then the large symmetry group of $X$ does the rest.)


So, here is the thing I can't explain. Out of curiosity, I computed the above determinant and factored it. It turns out that
$$\det \begin{pmatrix}
0 & 0 & 3 r^2 & 3 s^2\\
3 r^2 & 3 s^2 & 6pr & 6 qs \\
6pr & 6 qs & 3 p^2 & 3 q^2 \\
3 p^2 & 3 q^2 & 0 & 0 \\
\end{pmatrix} = - 81 \det\begin{pmatrix} p & q \\ r & s \end{pmatrix}^4 .$$


Is there any deep reason for this? Does that determinant even have any significance away from the $18$ points which describe lines on $X$?



A funny factorization of the Jacobian coming from the lines on the Fermat cubic
The Jacobian matrix consists of coefficients of $t^3,t^2u,tu^2,u^3$ in the following $4$ partial derivatives $$\partial\_p F(t,u,pt+ru,qt+su) = \partial\_{y}F(t,u,pt+ru,qt+su)t,\\
\ldots\\
\partial\_s F(t,u,pt+ru,qt+su)= \partial\_{z}F(t,u,pt+ru,qt+su)u.$$ 


Following B. Wellington's comment, (up to permutation of columns) this matrix is by definition the Sylvester matrix of the dehomogenizations $P(t)$ and $Q(t)$ of $P(t,u) = \partial\_{y}F(t,u,pt+ru,qt+su)$ and $Q(t,u) = \partial\_{z}F(t,u,pt+ru,qt+su)$ respectively, regarded as homogenous polynomials in two variables $t$ and $u$. 


In our particular case, $P(t,u) = 3(pt + ru)^2$ and $Q(t,u) = 3(qt+su)^2$. Since the resultant of two polynomials is defined as the product of the differences of their roots,
it follows immediately that the determinant of the Jacobian matrix (up to sign) is
$$Res(P(t),Q(t)) = 81 Res(pt+r,qt+s)^4 = 81(pq-rs)^4.$$


