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.
 A: Even without thinking about it, I can tell you that the answer is almost certainly no for $m \ge 3$. Indeed, this would in contradiction with Fermat's last theorem, if you let $a = x+z$, $b=y+z$, and $c=x+y+z$.
Now it's possible that the implication is still true, because it's possible that there does not exist any $x,y,z$ such that $(x+y)^m = x^m+y^m+z^m$. But a computer helped me find $(1+8)^3 = 1^3 + 8^3 + 6^3$, however $(1+8+6)^3 = 3375$ whereas $(1+6)^3+(1+8)^3 = 3087$.
A: Expanding the given condition using binomial we get,
$$ z^m = {m\choose{1}} xy^{m-1} + {m\choose{2}} x^2y^{m-2} +...........+{m\choose{m-1}} x^{m-1}y $$
Now if we expand the claim,  we get,
$$
\sum \frac{m!}{a!b!c!} x^ay^bz^c = z^m + \Bigg( {m\choose{1}} xz^{m-1} + {m\choose{2}} x^2z^{m-2} +...........+{m\choose{m-1}} x^{m-1}z \Bigg)  + \Bigg( {m\choose{1}} yz^{m-1} + {m\choose{2}} y^2z^{m-2} +...........+{m\choose{m-1}} y^{m-1}z \Bigg)
$$ 
where $a, b, c$ are natural numbers such that $a+b+c = m$ and $a,b,c \neq m$.
We already know the value of $z^m$ so we can substitute that, 
$$
\sum \frac{m!}{a!b!c!} x^ay^bz^c = \Bigg(  {m\choose{1}} xy^{m-1} + {m\choose{2}} x^2y^{m-2} +...........+{m\choose{m-1}} x^{m-1}y  \Bigg) + \Bigg( {m\choose{1}} xz^{m-1} + {m\choose{2}} x^2z^{m-2} +...........+{m\choose{m-1}} x^{m-1}z \Bigg)  + \Bigg( {m\choose{1}} yz^{m-1} + {m\choose{2}} y^2z^{m-2} +...........+{m\choose{m-1}} y^{m-1}z \Bigg)
$$ 
We can simplify this to, 
$$
\sum \frac{m!}{a!b!c!} x^ay^bz^c = \Bigg(  {m\choose{1}} ( xy^{m-1} + yz^{m-1} + zx^{m-1} ) +
 {m\choose{2}} (x^2y^{m-2} + y^2z^{m-2} + z^2x^{m-2})...........+{m\choose{m-1}}( x^{m-1}y^ + y^{m-1}z + z^{m-1}x ) \Bigg) 
$$ 
I can't see how to prove that $LHS=RHS$ from here.  In fact, since you are not sure whether your claim was true, this seems to indicate that the claim is false.
Can anyone try something from here?
Edit As the other answer says the claim is clearly contradicting FMT.
A: 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
$$
z^m=(x+y)^m-x^m-y^m,
$$
$$
z^m=\sum_{k=1}^{m-1}\binom{m}{k}x^{m-k}y^k.
$$
$$
\left(\frac{z}{x}\right)^m=\sum_{k=1}^{m-1}\binom{m}{k}\left(\frac{y}{x}\right)^k...
$$
