（测试帖）如何判断这族多项式的可约性

$$
\begin{aligned}
R_n(x)&=(x-1)\left[(3-x)^n-(1-x)^n\right]+\left(x^2-3x+3 \right)^n-1\\\\
r_n(x)&=\begin{cases}
\dfrac{R_n(x)}{x(x-1)(x-2)(x^2-4x+5)}&n\equiv0\pmod4\\\\
\dfrac{R_n(x)}{x(x-1)(x-2)}&n\equiv2\pmod4\\\\
\dfrac{R_n(x)}{x(x-1)}&n\equiv1\pmod2
\end{cases}
\end{aligned}
$$

求证：$n\ge2$ 时，$r_n$ 在 $\mathbb{Q}$ 上不可约

mma验证代码：

R=(x-1)((3-x)^n-(1-x)^n)+(x^2-3x+3)^n-1;
r=Piecewise[{{R/(x(x-1)),Mod[n,2]==1},{R/(x(x-1)(x-2)),Mod[n,4]==2},{R/(x(x-1)(x-2)(x^2-4x+5)),Mod[n,4]==0}}];
Table[Factor[r],{n,1,10}]//IrreduciblePolynomialQ