1. **Stating the problem:** Simplify the expression $$\left( \frac{a+b}{ab^2} - \frac{a-b}{a^2b} \right)^2 \cdot \frac{a^5b^6}{a^5b^2 - ab^6} - \frac{2ab}{a^2 - b^2}$$ 2. **Formula and rules:** - To subtract fractions, find a common denominator. - Factor expressions where possible. - Use difference of squares: $a^2 - b^2 = (a-b)(a+b)$. - Simplify powers and cancel common factors carefully. 3. **Simplify the difference inside the square:** $$\frac{a+b}{ab^2} - \frac{a-b}{a^2b} = \frac{(a+b) \cdot a - (a-b) \cdot b}{a^2 b^2}$$ Calculate numerator: $$a(a+b) - b(a-b) = a^2 + ab - ab + b^2 = a^2 + b^2$$ So, $$\frac{a+b}{ab^2} - \frac{a-b}{a^2b} = \frac{a^2 + b^2}{a^2 b^2}$$ 4. **Square the fraction:** $$\left( \frac{a^2 + b^2}{a^2 b^2} \right)^2 = \frac{(a^2 + b^2)^2}{a^4 b^4}$$ 5. **Simplify the denominator in the big fraction:** $$a^5 b^2 - a b^6 = a b^2 (a^4 - b^4)$$ Factor $a^4 - b^4$ as difference of squares: $$a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)$$ So denominator is: $$a b^2 (a^2 - b^2)(a^2 + b^2)$$ 6. **Rewrite the big fraction:** $$\frac{a^5 b^6}{a^5 b^2 - a b^6} = \frac{a^5 b^6}{a b^2 (a^2 - b^2)(a^2 + b^2)}$$ Cancel common factors: $$= \frac{\cancel{a^5} b^6}{a b^2 (a^2 - b^2)(a^2 + b^2)} = \frac{a^{4} b^{6}}{a b^{2} (a^{2} - b^{2})(a^{2} + b^{2})}$$ Cancel $a$ and $b^2$: $$= \frac{a^{4-1} b^{6-2}}{(a^{2} - b^{2})(a^{2} + b^{2})} = \frac{a^{3} b^{4}}{(a^{2} - b^{2})(a^{2} + b^{2})}$$ 7. **Multiply the squared fraction by this simplified fraction:** $$\frac{(a^{2} + b^{2})^{2}}{a^{4} b^{4}} \cdot \frac{a^{3} b^{4}}{(a^{2} - b^{2})(a^{2} + b^{2})} = \frac{(a^{2} + b^{2})^{2} a^{3} b^{4}}{a^{4} b^{4} (a^{2} - b^{2})(a^{2} + b^{2})}$$ Cancel $b^{4}$: $$= \frac{(a^{2} + b^{2})^{2} a^{3}}{a^{4} (a^{2} - b^{2})(a^{2} + b^{2})}$$ Cancel $a^{3}$ with $a^{4}$: $$= \frac{(a^{2} + b^{2})^{2}}{\cancel{a^{4}} (a^{2} - b^{2})(a^{2} + b^{2})} \cdot \frac{a^{3}}{\cancel{a^{3}}} = \frac{a (a^{2} + b^{2})^{2}}{a^{4} (a^{2} - b^{2})(a^{2} + b^{2})}$$ More precisely: $$= \frac{(a^{2} + b^{2})^{2} a^{3}}{a^{4} (a^{2} - b^{2})(a^{2} + b^{2})} = \frac{a^{3} (a^{2} + b^{2})^{2}}{a^{4} (a^{2} - b^{2})(a^{2} + b^{2})}$$ Cancel $a^{3}$ with $a^{4}$ leaving $a$ in denominator: $$= \frac{(a^{2} + b^{2})^{2}}{a (a^{2} - b^{2})(a^{2} + b^{2})}$$ Cancel one $(a^{2} + b^{2})$: $$= \frac{a^{2} + b^{2}}{a (a^{2} - b^{2})}$$ 8. **Now subtract the last term:** $$\frac{a^{2} + b^{2}}{a (a^{2} - b^{2})} - \frac{2ab}{a^{2} - b^{2}}$$ Rewrite second fraction with denominator $a (a^{2} - b^{2})$: $$\frac{2ab}{a^{2} - b^{2}} = \frac{2ab \cdot a}{a (a^{2} - b^{2})} = \frac{2a^{2} b}{a (a^{2} - b^{2})}$$ 9. **Combine the two fractions:** $$\frac{a^{2} + b^{2}}{a (a^{2} - b^{2})} - \frac{2a^{2} b}{a (a^{2} - b^{2})} = \frac{a^{2} + b^{2} - 2a^{2} b}{a (a^{2} - b^{2})}$$ 10. **Simplify numerator:** $$a^{2} + b^{2} - 2a^{2} b$$ No further factorization is obvious, so this is the simplified form. **Final answer:** $$\boxed{\frac{a^{2} + b^{2} - 2a^{2} b}{a (a^{2} - b^{2})}}$$