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?

THEOREM. 
The equation 
$$
aqx^p+x^q=1\tag 1
$$
admits root $x$ such that 
$$
x^n=\frac{n}{q}\sum^{\infty}_{k=0}\frac{\Gamma(\{n+pk\}/q)(-q a)^k}{\Gamma(\{n+pk\}/q-k+1)k!}\textrm{, }n>0\tag 2
$$
where $\Gamma(x)$ is Euler's the Gamma function. (Note that the root of $(1)$, for no confusion is given from $(2)$ with $n=1$).
Here the sum $(2)$ is valid for all real numbers $n,p$ and $q$ and for complex $a$ with
$$
|a|\leq |p|^{p/q}|p-q|^{(p-q)/q}.\tag 3
$$
$\ldots$ You want to solve the equation 
$$
(1+\xi)^m(1-\xi)^n=C.\tag 4
$$ 
Set $1+\xi=x$, then $(4)$ becomes $x^m(1-(x-1))^n=C$ or $x^m(2-x)^n=C$. Hence $x^{m/n}(2-x)=C^{1/m}$ or $2-x=C^{1/m}x^{-m/n}$. Hence $2x^{-1}-1=C^{1/m}x^{-m/n-1}$. Setting  $2x^{-1}=y$ we have $y-1=\frac{C^{1/m}}{2^{m/n+1}}y^{m/n+1}$ 
or 
$$
-C_1y^A+y=1,\tag 5
$$
where $C_1=-C^{1/m}2^{-m/n-1}$, $A=m/n+1$ and 
$$
\xi=2/y-1.\tag 6
$$
Hence using $(1),(2)$ we can extract a solution of $(5)$ and hence of $(4)$. 
