How to solve $B = x^c - (1 - x)^c$ How to solve for x ? Where we are interested in the range  $0 < x < 1$ and $C \neq 0$.    
$$ B = x^c - (1 - x)^c.$$
The only thing I could come up with is to substitute 
      $$ x = \sin^2y $$ 
but I could not get anywhere.
    $$ B = (\sin^2y)^C - (1 - \sin^2y)^C $$
    $$ B = (\sin^2y)^C - (\cos^2y)^C  $$
    $$ B = (\sin y)^{2C} - (\cos y)^{2C}$$
Edit: I guess we will have to calculate values.
Example using Java:
public class Calculate {
public static void main(String args[]){ 
    for(double c = 0.5; c < 3; c += 0.5){
        for(double x = 0.5; x < 1; x += 0.1){
            double b = Math.pow(x, c) - Math.pow(1 - x, c);
            System.out.format(" x= %.2f c= %.2f b= %.3f %n", x, c, b);
        }
    }   
}   

Edit2:
$$ x^z - (1 - x)^z $$

http://www.wolframalpha.com/input/?i=+x%5Ez+-+%281+-+x%29%5Ez 
Edit3:
Given C and B, this code finds x.
import java.util.Scanner;
public class Z{
    public static void main(String args[]){  
        double c, b;
        double x = 0.000001;
        boolean outB = false;       
    Scanner input = new Scanner(System.in);
    System.out.println("Enter a value for C");
    c = input.nextDouble();
    System.out.println("Enter a value for B");
    b = input.nextDouble(); 
    while((b - Math.pow(x, c) + Math.pow(1 - x, c)) > 0.00001){
        x += 0.00001;
        if(x <= 0 || x >= 1){
            outB = true;
            break;
        }       
    }
    if(outB){
        System.out.println("Out of bounds!");
    } else {
        System.out.format(" x= %.3f c= %.3f b= %.3f %n", x, c, b);
    }   
}
}   

 A: For "most" values of $c$ and $B$, the Galois group of $x^c-(1-x)^c-B$ is not solvable, so in particular the equation $x^c-(1-x)^c=B$ can not (in general) be solved in radicals.
For a concrete example (suggested already in the comments), with $c=5$ and $B=2$, the Galois group is $S_5$. This is also the case for 
$$(c,B) \in \{ (5,3), (5,4), (5,5), (6,2), \ldots \}$$
and many many other examples.
A: For small $B$, you can use a series expansion:
$$ \eqalign{x =& \frac12+\frac14\,{\frac {{2}^{c}B}{c}}-{\frac {{2}^{3\,c} \left( c-1 \right) 
 \left( c-2 \right) {B}^{3}}{96\,{c}^{3}}}+{\frac {{2}^{5\,c} \left( c
-1 \right)  \left( c-2 \right)  \left( 9\,{c}^{2}-23\,c+8 \right) {B}^
{5}}{7680\,{c}^{5}}}\cr&-{\frac {{2}^{7\,c} \left( c-1 \right)  \left( c-2
 \right)  \left( 225\,{c}^{4}-1138\,{c}^{3}+1799\,{c}^{2}-902\,c+136
 \right) {B}^{7}}{1290240\,{c}^{7}}}\cr&+{\frac {{2}^{9\,c} \left( c-1
 \right)  \left( c-2 \right)  \left( 3\,c-1 \right)  \left( 3675\,{c}^
{5}-26498\,{c}^{4}+69153\,{c}^{3}-76634\,{c}^{2}+31752\,c-3968
 \right) {B}^{9}}{371589120\,{c}^{9}}}\cr &+ \ldots}
$$
A: A picture says more than a thousand words. Below is a sketch of the curves $\;\color{red}{y=x^c}$ , the curves $\;\color{green}{y=(1-x)^c}\;$ and their difference $\;y=x^c - y=(1-x)^c$ . The symmetries are clearly seen.The range of $\,c\,$ has been chosen as $\;-2 < c < +2\;$ with steps of $1/10$ . For $\;c < 0$ , the curves $\;\color{red}{y=x^c}\;$ as well as the curves $\;\color{green}{y=(1-x)^c}\;$ are above the line $\,y=1$ , for $\;c > 0\;$ they
are below that line and $\;0 \le y \le 1$ . A sample line $\;y=B\;$ is drawn for
$\;B=-0.7$ . The intersects with the black curves are solutions. I see no other way than numerically solve these equations, given some definite values of $\,B\,$ and $\,c$ . The picture might help then to find initial iterates.

