证明：证法一：在（-∞，+∞）上任取x1，x2且x1＜x2则f（x2）-f（x1）=x13-x23=（x1-x2）（x12+x1x2+x22）∵x1＜x2，∴x1-x2＜0．当x1x2＜0时，有x12+x1x2+x22=（x1+x2）2-x1x2＞0；当x1x2≥0时，有x12+x1x2+x22＞0...