如何求解一元三次方程3x^3-8x^2+6x-1=0 请写出具体过程

因式分解3x³-3-8x²+6x+2=03(x³-1)-2(4x²-3x-1)=03(x-1)(x²+x+1)-2(4x+1)(x-1)=0(x-1)(3x²+3x+3-8x-2)=0(x-1)(3x²-5x+1)=0x-1=0,3x²-5x+1=0x=1,x=(5-√13)/2,x=(5+√13)/2...