x^3-7x^2+6=0的三个根是多少?3Q

x^3-7x^2+6=0 x^3-7x^2+6x-6(x-1)=0 x(x-1)(x-6)-6(x-1)=0 (x-1)[x(x-6)-6]=0 (x-1)(x^2-6x-6)=0 (x-1)(x-3)(x+2)=0 x=1或者3或者-2