I suspect that you are intended to find a function whose graph looks exactly line y= f(x) but is translated (one translates geometric things- points and sets of points like graphs- not functions) by <-1,-1>. In particular, that means that, since f(0)= 0, we want g(-1)= -1. First, we want x= -1, in g, to "act like" x=0 in f. That is
g(x)= f(u) for some u so that when x= -1, u= 0: okay the simplest possible thing is u= x+1. If we write g(x)= f(x+1)= (x+1)3- 3(x+1)2, we have g(-1)= f(0) but we are not quite done: g(-1)= f(0)= 0 and we want g(-1)= f(0)-1. Fine: just subtract 1 from what we just got:
g(x)= (x+1)3- 3(x+1)2- 1.
In order to get it in the form "ax3+ b2+ cx+ d", you will need to multiply it out.