General solution or approximate solution Is there a known general or approximate explicit solution for $\xi$ in 
$$(1+\xi)^m (1-\xi)^n = C$$
 where $m$ and $n$ positive fractions and $C$ being constant?

Since $m$ and $n$ are fractions, you can take the LCM of their
denominators and write
$$
\left\{ \matrix{
  m = {p \over L}\quad n = {q \over L}\quad \left| {\;p,q,L \in N} \right. \hfill \cr 
  \left( {1 + x} \right)^{\,p} \left( {1 - x} \right)^{\,q}  = C^{\,L}  \hfill \cr}  \right.
$$
so that we can always reduce the problem to integral powers.
The approach I deem might work better is the following
Let's change the sign of the second term, so as to have a monic polynomial
$$ \bbox[lightyellow] {  
\left( {x + 1} \right)^{\,p} \left( {x - 1} \right)^{\,q}  = \left( { - 1} \right)^{\,q} C^{\,L}  = c\quad \left| {\;p,q,L \in N} \right.
}\tag{1}$$
Let's then consider that
$$
\left( {x + 1} \right)^{\,p} \left( {x - 1} \right)^{\,q}  = \left\{ {\matrix{
   {\left( {x^{\,2}  - 1} \right)^{\,p} \left( {x - 1} \right)^{\,q - p} } & {0 \le q - p}  \cr 
   {\left( {x^{\,2}  - 1} \right)^{\,q} \left( {x + 1} \right)^{\,p - q} } & {0 < p - q}  \cr 
 } } \right.
$$
We can therefore concentrate to examine
$$
\left\{ \matrix{
  f(x,y) = \left( {x^{\,2}  - 1} \right)^{\,s} \left( {y - 1} \right)^{\,t}  = c \hfill \cr 
  y = x \hfill \cr}  \right.\quad \left| {\;s,t \in N} \right.
$$
that is
$$
\left\{ \matrix{
  y = 1 + \left( {{c \over {\left( {x^{\,2}  - 1} \right)^{\,s} }}} \right)^{\,1\,/\,t}  \hfill \cr 
  y = x \hfill \cr}  \right.
$$
where it is easy to define the domain of existence according to the sign
of $x^2-1,\; s, \; c, \; t$, and a sketch surely helps to focus the situation.
For example

The sketch clearly indicates that in this case (and for other $s$ and $t$ with same parity)
we will have only one solution for positive $c$, and one or three for negative $c$.
It also tells us how we can compute the solution(s) by any of the classical
methods of iterative approximations, and where it is appropriate to place the starting point.  
