设x1,x2属于R,且x1＜x2则f（x1）-f（x2）=x1^3-x2^3=(x1-x2）（x1^2+x1x2+x2^2)=(x1-x2）（x1^2+x1x2+1/4x2^2+3/4x2^2)=(x1-x2）[(x1+1/2x2)^2+3/4x2^2]由x1＜x2则x1-x2＜0又由[(x1+1/2x2)^2+3/4x2^2]＞0故(x1-x2）[...