Partition Of Graph's edges Into 3 Groups 
Let $G = (V, E)$ be a bipartite graph. 
Prove that there is a partition of the set of edges $E$ into 3 disjoint parts:
$E = E1 ∪ E2 ∪ E3$, $E1 ∩ E2 = E2 ∩ E3 = E3 ∩ E1 = ∅$, so that for
  every vertex $v$ of $G$ and for each $1 ≤ i ≤ 3$, the degree $deg_i(v)$ of $v$ in the graph $(V, E_i)$ satisfies:

$\lfloor{\frac{deg(v)}{3}}\rfloor$ $≤ d_i(v) ≤$ $\lceil{\frac{deg(v)}{3}}\rceil$, 

Where $deg(v)$ is the degree of $v$ in $G$.
(Hint: Split vertices to have maximum degree 3, and find a proper edge coloring by 3 colors.)

I didn't understand the hint , plus , even if I do have a graph of maximum degree 3 , according to Vizing it can have chromatic index(optimal edge coloring) of 4.
I'll be happy if someone could give me a better hint, since I am really lost. When I tried to solve this I haven't even thought about edge coloring...
Also, can this result be generalized?
 A: In general, if $G$ has maximum degree $\Delta$, then the chromatic index of $G$ is either $\Delta$ or $\Delta+1$. For some special graphs, like bipartite graphs, the chromatic index is always $\Delta$.
The result can be generalized as follows.
Let $G=(V,E)$ be a bipartite graph. Let $k\geq1$ be an integer.
Then $E$ can be partitioned into $k$ parts $E_1,E_2,...,E_k$ so that $\lfloor\dfrac{deg(v)}{k}\rfloor\leq{d_i(v)}\leq\lceil\dfrac{deg(v)}{k}\rceil$ for all $v\in{V}$, for all $i=1,...,k$.
As in the hint, we first split each vertex $v$ into $\lceil\dfrac{deg(v)}{k}\rceil$ vertices, where $\lfloor\dfrac{deg(v)}{k}\rfloor$ of them have degree $k$ and one vertex (if necessary) of degree $n-\lfloor\dfrac{deg(v)}{k}\rfloor{k}$.
Then the resulting graph $G'$ has maximum degree $k$, so the chromatic index of $G'$ is $k$, i.e. $E(G')$ is a union of $k$ matchings $M_1,...,M_k$.
Identify the points we split before, we get $E_1,...,E_k$, which is a partition of $E$. It remains to check that for each $i$, $d_i(v)$ satisfies the inequality for all $v\in{V}$.
Since $M_i$ is a matching, there is at most one edge in $M_i$ incident with each of the $\lceil\dfrac{deg(v)}{k}\rceil$ vertices corresponding to $v\in{G}$, we get ${d_i(v)}\leq\lceil\dfrac{deg(v)}{k}\rceil$.
Among those vertices in $G'$ corresponding to $v$, $\lfloor\dfrac{deg(v)}{k}\rfloor$ of them have degree $k$. Every such vertex must incident with an edge in $M_i$. Thus, $\lfloor\dfrac{deg(v)}{k}\rfloor\leq{d_i(v)}$.
