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Given 4 points as points = 0, 4, 1, 0, 2, -1, 3, 0;. I want to find the coefficients of f[x_] := a x^3 + b x^2 + c x + d; and its plot.

How to solve this in Mathematica easily?

You can also use LinearSolve

A = CoefficientArrays[f@points[[All, 1]], a, b, c, d] //Last;
B = points[[All, 2]];

$A=left(
beginarraycccc
0 & 0 & 0 & 1 \
1 & 1 & 1 & 1 \
8 & 4 & 2 & 1 \
27 & 9 & 3 & 1 \
endarray
right) qquad B=left(
beginarrayc
4 \
0 \
-1 \
0 \
endarray
right)$

LinearSolve[A, B]

-(1/6), 2, -(35/6), 4

Four points define the polynomial unambiguously, not fitting is necesary, so for an exact solution I would do a system of equations and use Solve

points = 0, 4, 1, 0, 2, -1, 3, 0
(* 0, 4, 1, 0, 2, -1, 3, 0 *)

f[x_] := a x^3 + b x^2 + c x + d


(f[#1] == #2) & @@@ points
(* 
 
 d == 4, 
 a + b + c + d == 0, 
 8 a + 4 b + 2 c + d == -1, 
 27 a + 9 b + 3 c + d == 0
 
*)


Solve[
 (f[#1] == #2) & @@@ points
 , a, b, c, d
 ]
(* a -> -(1/6), b -> 2, c -> -(35/6), d -> 4 *)

fit = FindFit[points, f[x], a, b, c, d, x]

a -> -0.166667, b -> 2., c -> -5.83333, d -> 4.

Plot[Evaluate[f[x] /. fit], x, 0, 5, 
 Epilog -> PointSize[Large], Red, Point@points]

To get exact results, you can use

Rationalize[fit] 

a -> -(1/6), b -> 2, c -> -(35/6), d -> 4

Alternatively, you can use Reduce or Solve (as in rhermans's answer) with alternative specification of the first argument:

ToRules @ Reduce[f /@ points[[All, 1]] == points[[All, 2]]] (* or *)
Solve[f /@ points[[All, 1]] == points[[All, 2]], a, b, c, d][[1]]

a -> -(1/6), b -> 2, c -> -(35/6), d -> 4

You can use InterpolatingPolynomial:

Expand @ InterpolatingPolynomial[
 0,4,1,0,2,-1,3,0,
 x
]

4 - (35 x)/6 + 2 x^2 - x^3/6