Final Answer
$\frac{\left(x^2+2xy+y^2\right)\left(2x^2-xy-y^2\right)}{\left(x^2-y^2\right)\left(x^2-xy-2y^2\right)}$
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Step-by-step Solution
Specify the solving method
1
Multiplying the fraction by $2x^2-xy-y^2$
$\frac{\frac{\left(x^2+2xy+y^2\right)\left(2x^2-xy-y^2\right)}{x^2-y^2}}{x^2-xy-2y^2}$
2
Divide fractions $\frac{\frac{\left(x^2+2xy+y^2\right)\left(2x^2-xy-y^2\right)}{x^2-y^2}}{x^2-xy-2y^2}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$
$\frac{\left(x^2+2xy+y^2\right)\left(2x^2-xy-y^2\right)}{\left(x^2-y^2\right)\left(x^2-xy-2y^2\right)}$
Final Answer
$\frac{\left(x^2+2xy+y^2\right)\left(2x^2-xy-y^2\right)}{\left(x^2-y^2\right)\left(x^2-xy-2y^2\right)}$