Let f(x)=ln(1−x1+x) . The set of values of 'α'  for which f(α)+f(α2)=f(αα2−α+1)  is satisfied are

f(α)+f(α)2=ln[(1−α1+α)(1−α21+α2)]=ln[(1−α)21+α2] f(αα2−α+1)=ln[1−αα2−α+11+αα2−α+1]=ln[(1−α)21+α2] ∴  f(α)+f(α2)=f(αα2−α+1)  for all values of α  for which the functions are defined, therefore1−α1+α>0⇒−1<α<1....(1) 1−α21+α2>0⇒1−α2>0⇒−1<α<1....(2) From (1) and (2), we have −1<α<1 ∴  The set of values ofα=(−1,1) .