Prove formula $\sum\limits_{i\ge 3}(6-i)f_i=12$ for graph Let $G$ be connected cubic graph without loops and multiple edges. $f_i$ $(i\ge 3)$ is a number of its faces such that every face bounded with $i$ edges. Prove that $$\sum\limits_{i\ge 3}(6-i)f_i=12$$ 
I don't have any clue about this problem. The only thing I know is that this system is right: 
\begin{cases}
p-q+r=2\\
3p=2q\\
ir \le 2q
\end{cases}  
 A: We have
$$\sum_{i \ge 3}(6-i) f_i = 6\sum_{i\ge 3}f_i - \sum_{i\ge 3} i f_i.$$
The first sum counts faces with $i$ sides for each possible $i$, so it's just counting faces, and simplifies to $r$. The second sum is equivalent to counting the edges on each face (since $i f_i$ counts the $i$ edges on each face with $i$ sides). It counts each edge twice, once for each face it borders, so it simplifies to $2q$. Therefore
$$\sum_{i \ge 3}(6-i) f_i = 6r - 2q.$$
From here, the calculation is straightforward. Since $p-q+r=2$, $r = q-p+2$, so $6r - 2q = 4q - 6p + 12$. This simplifies to just $12$ because $4q - 6p = 2(2q - 3p) = 0$.
A: Just for my sake, i will use the normal notation: $v=p,e=q,f=r.$
From $v-e+f=2$ we get $2v-2e+2f=4$ and the lhs can be changed to $2v-3v+2f=2f-v,$ because $3v=2e$. The later is given by the handshake lemma $$\sum _{v\in V} d_v=2e,$$ where $d_v$ is the degree of the vertex, but we know it is cubic hence $d_v=3.$
From $2f-v=4$ we obtain $2f-v+2=6$ and hence $2(2f-v+2)=12,$ but $2=v-e+f$ hence $$2(2f-v+v-e+f)=2(3f-e)=6f-2e=12.$$
Notice that $$f=\sum _{i\geq 3}f_i,$$ and $$2e=\sum _{i\geq 3}f_i,$$
why? 
