问道数学题一元三次方程k^3-3k^2+k+1= 0的根怎么求数学一元三次方程

分解因式（k^2-2k-1)(k-1)=0
k=1+2^0.5, 1- 2^0.5,1

k³-3k²+k+1= 0k(k²-2k+1)-k²+1=0k(k-1)²-(k²-1)=0k(k-1)²-(k-1)(k+1)=0(k-1)(k²-k-k-1)=0(k-1)(k²-2k-1)=0 k=1,k²-2k-1=0(k-1)²=2k-1=+,-√2k=1+,-√2k=1,1...