设f(x)=ax^3+bx^2+cx+d.函数过原点,故d=0. 则导函数f’(x)=3ax^2+2bx+c. ∵x=1,x=3时有极值, ∴3a+2b+c=0,27a+6b+c=0. 当X=3时有极小值0,说明点（3,0）在f(x)上,即27a+9b+3c=0. 解上述方程求得a、b、c、d的值即可.(...