7x^3-3x^2-3x-1=0...这种三次的要怎么解
问题描述：
7x^3-3x^2-3x-1=0...这种三次的要怎么解
答
x³-1+6x³-3x-3=0
(x-1)(x²+x+1)+3(2x²-x-1)=0
(x-1)(x²+x+1)+3(2x+1)(x-1)=0
(x-1)(x²+x+1+6x+3)=0
(x-1)(x²+7x+4)=0
x-1=0,x²+7x+4=0
x=1,x=(-7-√33)/2,x=(-7+√33)/2