1. **State the problem:** Simplify the expression $$\left( \frac{1}{x^2 + xy} + \frac{2}{x - y} - \frac{1}{x^2 - xy} \right) \cdot \frac{x^3 - 2x^2 y + xy^2}{x^2 + xy - y}$$ 2. **Factor where possible:** - Factor denominators: $$x^2 + xy = x(x + y)$$ $$x^2 - xy = x(x - y)$$ $$x^2 + xy - y = y(x + y)$$ - Factor numerator of second fraction: $$x^3 - 2x^2 y + xy^2 = x(x^2 - 2xy + y^2) = x(x - y)^2$$ 3. **Rewrite the expression with factored terms:** $$\left( \frac{1}{x(x + y)} + \frac{2}{x - y} - \frac{1}{x(x - y)} \right) \cdot \frac{x(x - y)^2}{y(x + y)}$$ 4. **Find common denominator inside the parentheses:** The denominators are $x(x + y)$, $x - y$, and $x(x - y)$. The least common denominator (LCD) is $x(x + y)(x - y)$. 5. **Rewrite each term with the LCD:** $$\frac{1}{x(x + y)} = \frac{\cancel{1} \cdot (x - y)}{x(x + y)(x - y)} = \frac{x - y}{x(x + y)(x - y)}$$ $$\frac{2}{x - y} = \frac{2 \cdot x(x + y)}{x(x + y)(x - y)} = \frac{2x(x + y)}{x(x + y)(x - y)}$$ $$\frac{1}{x(x - y)} = \frac{\cancel{1} \cdot (x + y)}{x(x + y)(x - y)} = \frac{x + y}{x(x + y)(x - y)}$$ 6. **Combine the fractions inside the parentheses:** $$\frac{x - y + 2x(x + y) - (x + y)}{x(x + y)(x - y)}$$ 7. **Simplify the numerator:** Expand $2x(x + y)$: $$2x^2 + 2xy$$ So numerator becomes: $$x - y + 2x^2 + 2xy - x - y = 2x^2 + 2xy - 2y$$ Simplify $x - x = 0$. 8. **Factor numerator:** $$2x^2 + 2xy - 2y = 2(x^2 + xy - y)$$ Recall from step 2 that $x^2 + xy - y = y(x + y)$, so: $$2y(x + y)$$ 9. **Rewrite the entire expression:** $$\frac{2y(x + y)}{x(x + y)(x - y)} \cdot \frac{x(x - y)^2}{y(x + y)}$$ 10. **Cancel common factors:** - Cancel $x$ numerator and denominator: $$\frac{2y(x + y)}{\cancel{x}(x + y)(x - y)} \cdot \frac{\cancel{x}(x - y)^2}{y(x + y)}$$ - Cancel $y$ numerator and denominator: $$\frac{2\cancel{y}(x + y)}{(x + y)(x - y)} \cdot \frac{(x - y)^2}{\cancel{y}(x + y)}$$ - Cancel $(x + y)$ numerator and denominator: $$\frac{2\cancel{(x + y)}}{\cancel{(x + y)}(x - y)} \cdot \frac{(x - y)^2}{(x + y)}$$ 11. **Multiply remaining terms:** $$\frac{2}{x - y} \cdot \frac{(x - y)^2}{x + y} = \frac{2(x - y)^2}{(x - y)(x + y)}$$ 12. **Cancel $(x - y)$:** $$\frac{2\cancel{(x - y)}(x - y)}{\cancel{(x - y)}(x + y)} = \frac{2(x - y)}{x + y}$$ **Final simplified expression:** $$\boxed{\frac{2(x - y)}{x + y}}$$