For x<0, f(x)= 1+x³. Therefore g(f(x))= g(1+x³)= (1+x³-1)^1/3= x.

for x> or =0, f(x)=x²-1. therefore g(f(x))= g(x²-1)=( x²-1+1)^1/2= x.

tore g(f(x))=x for all x belonging to (-infinity,infinity).

Thanks,

Saineel.