Proving the existence of a unique planar embedding 
Show that there is a unique planar embedding in which each vertex has degree 4 and each face has degree 3.

It is easy to just draw such a planar graph, but how to show the embedding is unique? The textbook doesn't have a solution.
 A: The Euler polyhedron formula $v+f=e+2$, together with $2e=4v$ and $2e=3f$ gives a unique solution $v=6, e=12, f=8$.
Fix a face and its vertices $a,b,c$. It has three neighbouring faces, giveing rise to three more veritces $d,e,f$. These must be pairwise different and different from $a,b,c$ as any conincidence would destroy one of the given conditions (check!).
So we have already found the $6$ vertices, and also $9$ of the edges. More edges can only be drawn among $d,e,f$ because $a.b,c$ have already degree $4$ 8now that we know that $d,e,f$ are distinct). So we must add all three edges $de,ef,fd$ and are done.
A: Use Euler's formula $F-E+V=2$
Note that each face has three edges, each edge borders two faces so that $E=\cfrac{3F}2$
Similarly each vertex meets four edges, each of which connects two  vertices so $E=\cfrac {4V}2=2V$
Substituting we fine $E=12, V=6, F=8$
This reduces the number of cases considerably, and the proof from there is not hard.
A: The problem comes from a textbook that just got done classifying the platonic graphs. You can simply argue that the graph in question is platonic and therefore it falls into the classification!
