But $ d(t) $ agrees with $ t^3 $ at $ t=1,2,3,4 $, yet is a cubic. So define $ p(t) = d(t) - t^3 $. Then $ p(t) $ is a polynomial of degree at most 3 (since $ d(t) $ is cubic, $ t^3 $ is cubic), and $ p(1) = p(2) = p(3) = p(4) = 0 $. - Abu Waleed Tea