If $G$ cubic planar graph with faces only $C_5$ and $C_6$, then $G$ has $12$ pentagons and $0$ hexagons

Question

I want to prove the questions in the title:

If $G$ is a connected cubic planar graph composed of only pentagons and hexagons, then it has $12$ pentagons and $0$ hexagons.

I was able to show it has $12$ pentagons, from Euler formula. However I am unable to proceed.

I believe it has to do with an inner structure it forms, because simply counting degrees and using Euler formula does not seem to work

Answer 1

The Truncated icosahedral graph has 12 pentagonal faces and 20 hexagonal faces. As noted in Wikipedia, it is connected, cubic, Hamiltonian, regular and, of course, it is planar. See also the MathWorld article.