∵f（x）=x3-x(x2+1)2，∴f′（x）=(3x2-1)(x2+1)2-(x3-x)•2•(x2+1)•2x(x2+1)4，令f′（x）=0，得（3x2-1）（x2+1）2-（x3-x）•2•（x2+1）•2x=0，∴（3x2-1）（x2+1）-4x2（x2-1）=0，整理，得x4-6x2+1=0，解...