Question 28
Find real number $a$ such that $x = \frac{1}{2}$ is a solution of the equation
$$
\left( \frac{x+1}{1-x} - \frac{1-x}{x+1} - \frac{4x^2}{x^2-1} \right) \left( \frac{1}{a^3+a^2} - \frac{1-a}{a^2} - 1 \right) = \frac{1}{2}
$$
$$
\left( \frac{x+1}{1-x} - \frac{1-x}{x+1} - \frac{4x^2}{x^2-1} \right) \left( \frac{1}{a^3+a^2} - \frac{1-a}{a^2} - 1 \right) = \frac{1}{2}
$$