Question

Помогите решить 2m^2-n^2/m^3+n^3 - m-n/m^2-mn+n^2

Answer 1

$\frac{2m^2-n^2}{m^3+n^3}-\frac{m-n}{m^2-mn+n^2};\\ (m^2-mn+n^2)(m+n)=m^3-m^2n+mn^2+m^2n-mn^2+n^3=\\ =m^3+n^3;\\ =\frac{2m^2-n^2-(m-n)(m+n)}{m^3+n^3}=\frac{2m^2-n^2+(n-m)(m+n)}{m^3+n^3}=\\ =\frac{2m^2-n^2+n^2-m^2}{m^3+n^3}=\frac{m^2}{m^3+n^3}$