由欧拉公式，\(x^n-1在复数域上的n个解为x_k=e^{i(\frac{2kΠ}{n})}\)，k=0,1,2,.....,n-1
其中\(x_k=cos\frac{2kΠ}{n}+isin\frac{2kΠ}{n}，k=0,1,2,.....,n-1\)
\(x^n-1\)在复数域上的标准分解为\(x^n-1=(x-x_0)(x-x_1).....(x-x_{n-1})\)
经观察，\((x-x_i)(x-x_{n-i}=x^2+1-2xcos\frac{2iΠ}{n}\)，为实数域上的既约多项式
当n为奇数时，\(x_0=1，则x^n-1=(x-1)[x^2+1-2xcos\frac{2Π}{n}][x^2+1-2xcos\frac{4Π}{n}].....[x^2+1-2xcos\frac{(n-1)Π}{n}]\)
当n为偶数时，\(x_0=1，x_{n/2}=-1，则x^n-1=(x-1)(x_1)[x^2+1-2xcos\frac{2Π}{n}][x^2+1-2xcos\frac{4Π}{n}].....[x^2+1-2xcos\frac{(n-2)Π}{n}]\)
原文：https://www.cnblogs.com/valar-morghulis/p/14012214.html