I deleted my previous answer because it was wrong!!!!!!!!!!!!!
First, let's do it the easy way (which I didn't see until I did it the long way.)
Subtract 2 from each of the ordinates: 0,1,8,27 ...this is obviously n^3; therefore the original ordinates are just n^3 +2.
If, like me, you didn't see the easy solution immediately, then make a difference table. You will see that the third differences are constant and equal to 6. This immediately tells you that the original function is a cubic and this is known from the study of finite differences (See "Calculus of Finite Differences" by Jordan).
The second differences are an arithmetic sequence, namely 6(n+1). This is always true when the 3rd differences are constant.
The first differences are given by 3n^2 + 3n + 1 and finally the original function is given by n^3 + 2 as previously stated.
The method of back tracking from the difference table takes longer to explain than I can go through here; suffice to say, the reference given above will show you how.
