f(x)=-x^3+m,
令x1<x2
f(x2)-f(x1)
=[-x2^3+m] - [-x1^3+m]
= x1^3-x2^3
=(x1-x2)(x1^2+x2^2+x1x2)
=(x1-x2) * [(x1+1/2 x2)^2 + 3/4 x2^2]
∵(x1+1/2 x2)^2 + 3/4 x2^2>0,x1-x2<0
∴f(x2)-f(x1)<0,f(x2)<f(x1)
∴函数在R上是减函数
∵函数是奇函数
∴f(-x)=-f(x)
∴-(-x)^3+m=-(-x^3+m),x^3+m=x^3-m
∴m=0