Let y=f(x)=x3-1,
x3=y+1,
x=∛(y+1).
If g(y)=x=∛(y+1), then g(x)=∛(x+1).
f(g(x))=g(f(x))=x so g(x)=f-1(x) (the inverse of f).
y=f(x): For every x there is a unique y;
y=g(x): For every y there is a unique x.
So the mapping is one-to-one.
so f-1(x)=∛(x+1).