令x=m^4-m^3*ny=n^3*m-n^4∴x-y=m^4-m^3*n-n^3*m+n^4=m(m^3-n^3)-n(m^3-n^3)=(m-n)(m^3-n^3)(1)当m=n时,x-y=0即：m^4-m^3*n=n^3*m-n^4(2)当m≠n时,x-y＞0即：m^4-m^3*n＞n^3*m-n^4综上所述,m^4-m^3*n≥n^3*m-n^4...