The most direct, but by no means the most efficient, way of finding the cubic polynomial is to start with
#f(x) = a_0+a_1 x+a_2 x^2 +a_3 x^3#
and substitute the coordinates to find the coefficients :
#a_0 +a_1 times 0 +a_2 times 0^2 + a_3 times 0^3 =135#
#a_0 +a_1 times 1 +a_2 times 1^2 + a_3 times 1^3 =156#
#a_0 +a_1 times 2 +a_2 times 2^2 + a_3 times 2^3 =115#
#a_0 +a_1 times 3 +a_2 times 3^2 + a_3 times 3^3 =0#
This means that we have to solve
# ((1,0,0,0),(1,1,1,1),(1,2,4,8),(1,3,9,27))((a_0),(a_1),(a_2),(a_3)) = ((135),(156),(115),(0))#
We can use any standard method to solve this. Gauss elimination is an efficient approach, in which we start with the augmented matrix
# ((1,0,0,0,|,135),(1,1,1,1,|,156),(1,2,4,8,|,115),(1,3,9,27,|,0))#
and carry out row reductions.
#R_2 to R_2-R_1, R_3 to R_3-R_1,R_4 to R_4-R_1#
# ((1,0,0,0,|,135),(0,1,1,1,|,21),(0,2,4,8,|,-20),(0,3,9,27,|,-135))#
#R_3 to R_3-2R_2,R_4 to R_4-3R_2#
# ((1,0,0,0,|,135),(0,1,1,1,|,21),(0,0,2,6,|,-62),(0,0,6,24,|,-198))#
#R_3 to R_3/2#
# ((1,0,0,0,|,135),(0,1,1,1,|,21),(0,0,1,3,|,-31),(0,0,6,24,|,-198))#
#R_4 to R_4-6R_3#
# ((1,0,0,0,|,135),(0,1,1,1,|,21),(0,0,1,3,|,-31),(0,0,0,6,|,-12))#
This allows us to find the coefficients
#a_3 = -2, a_2 = -25, a_1=48, a_0=135#
So
#f(x) = 135+48x-25x^2-2x^3#
We could have found this by more advanced methods like Newton interpolation or Lagrange interpolation so that the calculation could be done more efficiently.
