Does $(x+y)^m=x^m+y^m+z^m$ imply $(x+y+z)^m=(x+z)^m+(y+z)^m$? Let $x,y,z,m\in\mathbb{N}$, and $x,y,z,m>0$, and also $x>y$. 
My problem is to understand if, under these sole hypotheses, we can prove that

$(x+y)^m=x^m+y^m+z^m \Longrightarrow (x+y+z)^m=(x+z)^m+(y+z)^m.$

If yes, how can we prove it? 
If not, which other hypotheses are needed, in order to make the implication true?
EDIT: I am also interested in the softer versions of the statement, i.e. whether we can prove or not that

$(x+y)^m=x^m+y^m+z^m \Longrightarrow (x+y+z)^m\lessgtr (x+z)^m+(y+z)^m,$

and, if not, which additional conditions we need to make the statement(s) true.

Here's my attempt, using the binomial expansion. About the LHS:
$$
x^m+y^m+z^m=(x+y)^m,
$$
$$
x^m+z^m+y^m+z^m=(x+y)^m+z^m,
$$
$$
(x+z)^m+(y+z)^m-\sum_{k=1}^{m-1}\binom{m}{k}z^{m-k}(x^k+y^k)=(x+y+z)^m-\sum_{k=1}^{m-1}\binom{m}{k}z^{m-k}(x+y)^k.
$$
So, here we can isolate the RHS. With the notation of @Najib, we find
$$
c^m-a^m-b^m=\sum_{k=1}^{m-1}\binom{m}{k}z^{m-k}[(x+y)^k-x^k-y^k].
$$
$$
\frac{c^m-a^m-b^m}{z^m}=\sum_{k=1}^{m-1}\binom{m}{k}\frac{(x+y)^k-x^k-y^k}{z^k}.
$$
On the other hand, the LHS can also be written
$$
z^m=(x+y)^m-x^m-y^m,
$$
$$
z^m=\sum_{k=1}^{m-1}\binom{m}{k}x^{m-k}y^k.
$$
$$
\left(\frac{z}{x}\right)^m=\sum_{k=1}^{m-1}\binom{m}{k}\left(\frac{y}{x}\right)^k...
$$
