Answer
$\dfrac{(m-n)^{2}}{m+n}\cdot\dfrac{m}{m^{2}-mn}=\dfrac{m-n}{m+n}$
Work Step by Step
$\dfrac{(m-n)^{2}}{m+n}\cdot\dfrac{m}{m^{2}-mn}$
Factor the denominator of the second fraction by taking out common factor $m$:
$\dfrac{(m-n)^{2}}{m+n}\cdot\dfrac{m}{m^{2}-mn}=\dfrac{(m-n)^{2}}{m+n}\cdot\dfrac{m}{m(m-n)}=...$
Multiply the two rational expressions and simplify by removing the repeated factors in the numerator and the denominator:
$...=\dfrac{m(m-n)^{2}}{m(m+n)(m-n)}=\dfrac{(m-n)^{2}}{(m+n)(m-n)}=\dfrac{m-n}{m+n}$
