Answer
$\dfrac{x^{2}+2xy+y^{2}}{x^{2}-y^{2}}\cdot\dfrac{2x^{2}-xy-y^{2}}{x^{2}-xy-2y^{2}}=\dfrac{2x+y}{x-2y}$
Work Step by Step
$\dfrac{x^{2}+2xy+y^{2}}{x^{2}-y^{2}}\cdot\dfrac{2x^{2}-xy-y^{2}}{x^{2}-xy-2y^{2}}$
Factor the numerator and denominator of the first fraction:
$\dfrac{(x+y)^{2}}{(x-y)(x+y)}\cdot\dfrac{2x^{2}-xy-y^{2}}{x^{2}-xy-2y^{2}}=...$
Simplify the first fraction
$...=\dfrac{(x+y)}{(x-y)}\cdot\dfrac{2x^{2}-xy-y^{2}}{x^{2}-xy-2y^{2}}=...$
Evaluate the product
$...=\dfrac{2x^{3}-x^{2}y-xy^{2}+2x^{2}y-xy^{2}-y^{3}}{x^{3}-x^{2}y-2xy^{2}-x^{2}y+xy^{2}+2y^{3}}=...$
Simplify like terms:
$...=\dfrac{2x^{3}+x^{2}y-2xy^{2}-y^{3}}{x^{3}-2x^{2}y-xy^{2}+2y^{3}}=...$
Factor the numerator and the denominator by grouping terms and simplify:
$...=\dfrac{(2x^{3}+x^{2}y)-(2xy^{2}+y^{3})}{(x^{3}-2x^{2}y)-(xy^{2}-2y^{3})}=\dfrac{x^{2}(2x+y)-y^{2}(2x+y)}{x^{2}(x-2y)-y^{2}(x-2y)}=$
$...=\dfrac{(x^{2}-y^{2})(2x+y)}{(x^{2}-y^{2})(x-2y)}=\dfrac{2x+y}{x-2y}$
