f'(x)=x^2-1,f'(x)=0,x=±1x>1时,f'(x)>0,(增函数）-10,（增函数）所以在x=1时,有极小值,f(1)=-2/3,解f(x)=-2/3,(1/3)x^3-x=-2/3,x^3-3x=-2,x^3-x+2=0,(x-1)(x^2+x-2)=0 (x-1)(x-1)(x+2)=0 得到x1=x2=1,x3=-2,所以应...