7年级数学题请验证下列等式是否成立(3^3+2^3)/(3^3+1^3)=(3+2)/(3+1) , (4^3+3^3)

经过验算,成立
(1)
(6^3+2^3)/(6^3+4^3)=(6+2)/(6+4)
(n^3+m^3)/[n^3+(n-m)^3]=(n+m)/[n+(n-m)]
(2)
(100^3+99^3)/(100^3+1^3)+(99^3+97^3)/(99^3+2^3)+(98^3+95^3)/(98^3+98^3+3^3)+...+(51^3+1^3)/(51^3+50^3)
=(100+99)/(100+1)+(99+97)/(99+2)+……+(51+1)/(51+50)
=(199+196+193+……+52)/101
=[(199+52)+(196+55)+……+(127+124)]/101
=251*25/101
=6275/101

验算化成数就可以了，简单，就不说了
(3^3+2^3)/(3^3+1^3)=(3+2)/(3+1)
把3看成m 把2看成n 就得到下面的公式
(m^3+n^3)/(m^3+(m-n)^3)=(m+n)/(m+m-n)
(m^3+n^3)/(m^3+(m-n)^3)=(m+n)(m^2+mn+n^2)/(m+m-n)(m^2-m(m-n)+(m-n)^2)

(m^3+n^3)/(m^3+(m-n)^3)=(m+n)/(m+m-n)
(m^3+n^3)/(m^3+(m-n)^3)=(m+n)(m^2+mn+n^2)/(m+m-n)(m^2-m(m-n)+(m-n)^2)
=(m+n)/(m+m-n)
（100^3+99^3)/(100^3+1^3）+（99^3+97^3)/(99^3+2^3)…………+(51^3+1^3)/(51^3+50^3)
=199/101+196/101+193/101……52/101
=6275/101