\begin{info}
\section{Results for the cubic $L_8$}
We consider the following cubic
\begin{equation}\tag{$L_8$}X_0^3+X_1^3+X_2^3+X_3^3=0
\end{equation}
\subsection{The Galois group}
The splitting Galois group for the
cubic is given by
\[G=\langle c^1\rangle.
\]
\subsection{The Picard group}The orbits of the 27 lines are given by
\[\begin{split}
O_{0}&=\{L(0,0)\},\\
O_{1}&=\{L(1,0),L(2,0)\},\\
O_{2}&=\{L(0,1),L(0,2)\},\\
O_{3}&=\{L(1,1),L(2,2)\},\\
O_{4}&=\{L(2,1),L(1,2)\},\\
O_{5}&=\{M(0,0),M(0,1)\},\\
O_{6}&=\{M(1,0),M(2,1)\},\\
O_{7}&=\{M(2,0),M(1,1)\},\\
O_{8}&=\{M(0,2)\},\\
O_{9}&=\{M(1,2),M(2,2)\},\\
O_{10}&=\{N(0,0),N(0,2)\},\\
O_{11}&=\{N(1,0),N(2,2)\},\\
O_{12}&=\{N(2,0),N(1,2)\},\\
O_{13}&=\{N(0,1)\},\\
O_{14}&=\{N(1,1),N(2,1)\}.
\end{split}\]
In order to describe
the Picard group, we consider the cubic over $\overline {Q}$
as the projective plane blown up in six points
so that the $27$ lines are given by
\begin{align*}
E_1&=L(0,0),&E_2&=L(1,0),&E_3&=L(2,0),\\
E_4&=M(0,0),&E_5&=M(1,0),&E_6&=M(2,0),\\
Q_1&=L(1,1),&Q_2&=L(2,1),&Q_3&=L(0,1),\\
Q_4&=M(2,2),&Q_5&=M(0,2),&Q_6&=M(1,2),\\
R_{1,2}&=L(1,2),&R_{2,3}&=L(2,2),&R_{3,1}&=L(0,2)\\
R_{4,5}&=M(0,1),&R_{5,6}&=M(1,1),&R_{6,4}&=M(2,1)\\
R_{1,4}&=N(2,1),&R_{1,5}&=N(0,1),&R_{1,6}&=N(1,1)\\
R_{2,4}&=N(0,2),&R_{2,5}&=N(1,2),&R_{2,6}&=N(2,2)\\
R_{3,4}&=N(1,0),&R_{3,5}&=N(2,0),&R_{3,6}&=N(0,0).
\end{align*}
where the $E_i$ are the exceptional lines,
the $R_{i,j}$ is the strict lifting of the line $(P_iP_j)$
and $Q_i$ the strict lifting of the conic going through all points
except $P_i$. In the basis given by the lifting
$\Lambda$ of line not going through any of the six points blown up
and by the $6$ exceptional lines $E_i$, the Picard group
of the variety over $\mathbf Q$ is given by the matrix
\[\begin{pmatrix}
0&0&-1&-1\\
1&0&0&0\\
0&1&0&0\\
0&1&0&0\\
0&0&0&1\\
0&0&1&-1\\
0&0&0&1\\
\end{pmatrix}
\]
The effective cone is generated by

\[\begin{split}
[O_{1}]&= e_{0},\\
[O_{2}]&= e_{1},\\
[O_{3}]&=-2e_{0}-e_{1}-2e_{2}-e_{3},\\
[O_{4}]&=-2e_{1}-2e_{2}-e_{3},\\
[O_{5}]&=-2e_{0}-e_{1}-2e_{2}-e_{3},\\
[O_{6}]&=-e_{2},\\
[O_{7}]&=-e_{3},\\
[O_{8}]&=-e_{2},\\
[O_{9}]&=-e_{0}-e_{1}-e_{2}-e_{3},\\
[O_{10}]&=-2e_{0}-2e_{1}-3e_{2}-e_{3},\\
[O_{11}]&=-e_{1}-e_{2}-e_{3},\\
[O_{12}]&=-e_{1}-e_{2}-e_{3},\\
[O_{13}]&=-e_{1}-2e_{2},\\
[O_{14}]&=-e_{0}-e_{2},\\
[O_{15}]&=-2e_{0}-e_{2}-e_{3}.\\
\noalign{and the canonical sheaf is given by}\omega_V^{-1}&=-e_{0}-e_{1}-2e_{2}-e_{3}.
\end{split}\]
The corresponding polyhedron may be described as
\[\begin{split}
X_1 &> 0,\\
X_2 &> 0,\\
X_1-2X_2 &> -1,\\
X_1+2X_2-X_3 &> -1,\\
-1/2X_3 &> -1,\\
-X_1-2X_2+1/2X_3 &> -1,\\
-2X_2+X_3 &> 0,\\
-X_1+1/2X_3 &> 0,\\
-X_1+2X_2-1/2X_3 &> -1.
\end{split}\]
with $ 1$ as coefficient.
Thus the constant $\alpha(V)$ is $ 7/18$.

\subsection{The bad places}
\par
At the bad place $3$
We use the following reduced form

\[X_0^3+X_1^3+X_2^3+X_3^3=0
\]

of type $(1,1,1,q)$. Thus
the density at $p$ is $ 2$
and the zeta factor at $p$ is $ 8/27$

\subsection{Experimental curve}
The following graph represents the numerical data

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\end{info}
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