Question

$\frac{ m^{2}+2mn+n^{2}}{ m^{2}-n ^{2} } * \frac{m^{3}-n ^{3} }{m ^{3}+2m ^{2}n+2mn+n ^{3} }$

Answer 1

$\frac{m^2+2mn+n^2}{m^2-n^2}* \frac{m^3-n^3}{m^3+2m^2n+2mn^2+n^3}= \frac{(m+n)^2}{(m-n)(m+n)}* \frac{(m-n)(m^2+mn+n^2)}{(m^3+n^3)+(2m^2n+2mn^2)} \\ = \frac{(m+n)}{(m-n)}* \frac{(m-n)(m^2+mn+n^2)}{(m+n)(m^2-mn+n^2)+2mn(m+n)}= \\ =\frac{(m+n)}{(m-n)}* \frac{(m-n)(m^2+mn+n^2)}{(m+n)(m^2-mn+n^2+2mn)}=\frac{(m+n)}{(m-n)}* \frac{(m-n)(m^2+mn+n^2)}{(m+n)(m^2+mn+n^2)}= \\ =\frac{m^2+mn+n^2}{m^2+mn+n^2}=1$