If G is a graph such that $\Delta(G) = 3$ then $\chi'(G) \le 4$ $\chi'(G)$ being the chromatic index of $G$, $\chi(G)$ its chromatic number and  $\Delta(G)$ its maximum degree.
Here's the solution that I'm given, but I don't understand it:

Apply Brooks' theorem to the line graph of G.

I see how $\chi'(G) = \chi(L(G))$.
But a graph G with $\Delta(G) = 3$ can obviously have a line graph such that $\Delta(L(G)) > 3$, take for example:


Additionally the line graph could turn out to be an odd cycle...
Any idea where to go from there?
 A: You don't need to worry about odd cycles, because they can be colored with three colors and you are allowed four. 
You need to justify three things:


*

*Show that the maximum possible degree is $4$ (not so bad, I think).

*Show that thine graph cannot be the complete graph on five vertices, i.e. that you will not get a clique (you can check by going backwards if you like).

*You don't care if you get an odd cycle.


The you can use Brook's theorem in all its glory.
For further information, this result is called Vizing's Theorem.
A: Suppose $\Delta(G) = 3$. We need to show that 4 colors suffice to color the edges of the graph.
Let $uv$ be an edge in $G$.  How many edges share a common endpoint with this edge? Since $\Delta=3$, at most two other edges besides $uv$ are incident to the vertex $u$, and similarly for $v$.  So, at most $2+2=4$ edges share a common endvertex with edge $uv$.  Thus, in the line graph of $G$, the vertex $uv$ has degree at most 4.  If this line graph $L(G)$ is not an odd cycle or a complete graph, by Brook's theorem its vertices (and hence the edges of $G$) can be 4-colored.  
If the line graph is an odd-cycle, its vertices can be 2- or 3- colored, hence 4 colors suffice.  If the line graph is the complete graph $K_r$, then consider the value of $r$.  If $r=3$, this $K_3$ is the line graph of either $K_3$ or a star $K_{1,3}$, and in either case 3 colors suffice.  If $r \ge 4$, then this $K_r$ arose from a $G$ that is necessarily a star $K_{1,r}$. (This is a basic result on line graphs: any clique in $L(G)$ containing 4 or more vertices is induced by a star in $G$.)  Since $\Delta(G) =3$, this star has exactly 3 edges; thus $r=3$ and three colors suffice again. 
In other words, there is no clique of size 4 or more in the line graph $L(G)$ if $\Delta(G)=3$. More generally, all cliques in $L(G)$ arise from either stars $K_{1,r}$ in $G$ (i.e. the set of edges of $G$ that share a common endvertex) or triangles in $G$. 
