一个高一的指数函数的题

(m-n)/[m^(1/3)-n^(1/3)] - (m+n)/[m^(1/3)+n^(1/3)]
因为
(x^3-y^3)/(x-y)=x^2+xy+y^2
(x^3+y^3)/(x+y)=x^2-xy+y^2

所以原式=[m^(1/3)]^2+m^(1/3)*n^(1/3)+[n^(1/3)]^2
-{[m^(1/3)]^2-m^(1/3)*n^(1/3)+[n^(1/3)]^2}
=2m^(1/3)*n^(1/3)

通分得{（m-n)[m^(1/3)+n^(1/3)]}/[m^(1/3)-n^(1/3)]-

{（m+n)[m^(1/3)-n^(1/3)]}/[m^(1/3)+n^(1/3)] =

[2mn^(1/3)-2nm^(1/3)]/[m^(2/3)-n^(2/3)]提出公因式
2m^(1/3)n^(1/3)得，

原式=2m^(1/3)n^(1/3)