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?
 A: Hint.
As  $-1 < \xi < 1$ consider $\xi = \cos x$ then following
$$
1-\cos x = 2\sin^2(\frac x2)\\
1+\cos x = 2\cos^2(\frac x2)
$$
we have
$$
\sin^{2n}(\frac x2)\cos^{2m}(\frac x2) = \frac{C}{2^{n+m}}
$$
now making considerations over the $\sin^{2n}(\frac x2)\cos^{2m}(\frac x2)$ extrema ...
A: 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.  
A: 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 $(4)$. 
A: First I thought, by deriving the general solution was $\xi=\frac{m-n}{m+n}$. 
But for a solution with given $C$, I'd first reformulate the problem to $(1+\xi)(1-\xi)^a=c'$ with $a=n/m>0$ and $c'=c^{\frac{1}{m}}$...
A: Let WLOG 
$$(m,n)\in\mathbb N^2\tag1$$
and
$$(1+\xi)^m(1-\xi)^n=C.\tag2$$
So
$$m\log(1+\xi)+n\log(1-\xi)=\log C,\quad x\in(-1,1).\tag3$$
Note that
$$\lim_{x\to\pm1}LHS(3) = -\infty.$$
Maximum of $LHS(3)$ achieves when
$$\dfrac m{1+\xi} -\dfrac n{1-\xi}=0,$$
and that gives
$$\xi_m=\dfrac{m-n}{m+n}, \quad \max LHS(2) = \left(\dfrac {2m}{m+n}\right)^m\left(\dfrac {2n}{m+n}\right)^n.$$
Thus, the solution of $(2)$ exists only if
$$C \in\left[0,\dfrac{2^{m+n}m^mn^n}{(m+n)^{m+n}}\right].$$
If $\xi\not = \xi_m,$ then there are two solutions of $(2),$ which can be calculated using the iteration formulas
$$\xi_0 = -1,\quad \xi_{i+1}=1-C(1+\xi_i)^{-\frac mn}$$
for the less root and 
$$\xi_0=1,\quad \xi_{i+1}=C(1-\xi_i)^{-\frac nm}-1$$
for the greater root.
