已知一个数列的前 $n$ 项和 $S_n$ 表示为 $S_n = an^3 + bn^2 + cn + d$，其中 $a$、$b$、$c$ 和 $d$ 为常数。已知 $S_1 = 5$，$S_2 = 20$，$S_3 = 57$ 且 $S_4 = 128$。求 $S_n$。

1. Formulate a system of linear equations based on the given values of $S_n$.

$S_1 = a(1)^3 + b(1)^2 + c(1) + d = 5 \implies a + b + c + d = 5$
$S_2 = a(2)^3 + b(2)^2 + c(2) + d = 20 \implies 8a + 4b + 2c + d = 20$
$S_3 = a(3)^3 + b(3)^2 + c(3) + d = 57 \implies 27a + 9b + 3c + d = 57$
$S_4 = a(4)^3 + b(4)^2 + c(4) + d = 128 \implies 64a + 16b + 4c + d = 128$

2. Solve the system of linear equations using a graphing calculator.

Using a graphing calculator to solve the system of 4 linear equations:
$a = 2$
$b = -1$
$c = 4$
$d = 0$

3. State the final expression for $S_n$.

$S_n = 2n^3 - n^2 + 4n$