一元三次方程求解,x^3-3x^2-9x-5=0 x^3-13x^2+36x=0

(1)∵x^3-3x^2-9x-5=（x^3+x^2）-（4x^2+9x+5）=x^2(x+1)-(x+1)(4x+5)=(x+1)(x^2-4x-5)=(x+1)^2(x-5)=0
∴x=-1,或x=5
(2)∵x^3-13x^2+36x=x(x^2-13x+36)=x(x-4)(x-9)=0
∴x=0或x=4或x=9