Euler, Grinberg,… who's next?

Given a cubic planar hamiltonian graph with $F$ faces. Let $a_k$ be the number of face of degree $k$ inside and $b_k$ outside the Hamilton cycle. We have the following:

  1. $\sum \limits_k \left(a_k+b_k\right)k = 6F-12$ (due to Euler; my attempt)
  2. $\sum \limits_k \left(a_k-b_k\right)(k-2)=0$ (Grinberg’s Theorem)

Are these two all of this type or are there more?

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Let’s restrict it further to bipartite, cubic, hamiltonian Graphs $G$ only made up of $4$- or $6$-faces. These graphs can build out of $6$ faces of degree $4$ and the rest of degree $6$. Let $F$ be the number of faces, $V$ the number of Vertices, $E$ the number of edges and $a_k$ be the number of face of degree $k$ inside and $b_k$ outside the Hamilton cycle. Starting from Euler, we get:
$$
F+V=E+2\\
F=\sum_{k\in \{4,6\}} a_k+b_k = E-V+2
$$
For $3$-regular graphs, we have $2E=3V$ and $a_4+b_4=6$ so
$$
6+a_6+b_6= V\left(\frac{3}2-1\right)+2 \\
2(a_6+b_6)= V-8\tag{1}
$$
By combining $(1)$ and
$$a_4+b_4=6\tag{2}$$ with Grinberg’s formula (divided by $2$) in the given case
$$
(a_4-b_4)+2(a_6-b_6)=0\tag{3}
$$
we arrive at

  1. $$
    a_4+2a_6=\frac12V-1 \tag{$\frac12[(1)+(2)+(3)]$}
    $$

I checked some examples and it seems to work. Tell me if there is anything wrong or unclear.

Depending on $\frac V2$ being even or odd there are $1,3,5$ or $(0,)2,4,6$ possible $4$-faces inside the HC, where I’m not sure if $0$ is possible…