Closed form of the sum of two binomial expansions I would like to know if there is a simple closed form for the following expression:
$(x+y)^n + (x+z)^n$
Expanding the above I get
$(y^n + z^n) + nx(y^{n-1}+z^{n-1}) + \frac{(n-1)n}{2}x^2(y^{n-2} + z^{n-2}) + \mathcal{O}(x^3) $,
but it isn't clear to me if this can factor into something of the form $(x+w)^n + u$ where $w$ and $u$ depend on $y$ and $z$ (in general I mean, it is fairly easy to get something of this form for $n=1$ or $n=2$). 

Maybe you are looking for a symmetric formulation like this
$$
\eqalign{
  & \left( {x + y} \right)^{\,n}  + \left( {x + z} \right)^{\,n}  =   \cr 
  &  = \left( {x + \left( {{{z + y} \over 2}} \right) - \left( {{{z - y} \over 2}} \right)} \right)^{\,n}
  + \left( {x + \left( {{{z + y} \over 2}} \right) + \left( {{{z - y} \over 2}} \right)} \right)^{\,n}  =   \cr 
  &  = \left( {x + \left( {{{z + y} \over 2}} \right)} \right)^{\,n}
 \left( {\left( {1 - \left( {{{z - y} \over {2x + y + z}}} \right)} \right)^{\,n}  + \left( {1 + \left( {{{z - y} \over {2x + y + z}}} \right)} \right)^{\,n} } \right) \cr} 
$$
