Simplify. a) \( \frac{a b-a^{2}}{b^{2}-a^{2}} \div \frac{a^{2}}{b^{2}+a b} \) b) \( \frac{3 x^{2}+27}{2 x+6} \times \frac{6 x-18}{x^{4}-81} \) c) \( \frac{2 x^{2}-6 x}{x^{2}-1} \div \frac{x-3}{x^{2}-x} \div \frac{1}{3 x+3} \) d) \( \frac{x^{2}+x-6}{3 x^{2}-12 x} \div \frac{x^{3}-2 x^{2}}{x^{2}-16} \times \frac{1}{x+4} \) e) \( \frac{a^{2} b-a b}{a^{3}+a^{2}} \div \frac{a^{2} b-2 a b+b}{a^{3}-a} \) f) \( \frac{a^{2}-2 a+1}{a^{2}+2 a+1} \times \frac{a^{2}-1}{a^{2}} \times \frac{a^{2}+a}{a^{2}-a} \)

Let’s dive into these simplifications and have some fun while unraveling these algebraic expressions! a) To simplify \( \frac{a b-a^{2}}{b^{2}-a^{2}} \div \frac{a^{2}}{b^{2}+a b} \), we start with the first expression. We can factor \( a(b-a) \) in the numerator and \( (b+a)(b-a) \) in the denominator, leading to \( \frac{a(b-a)}{(b+a)(b-a)} \). The terms \( (b-a) \) cancel, giving us \( \frac{a}{b+a} \). Now, we flip the second fraction and multiply: \( \frac{a}{b+a} \cdot \frac{b^{2}+ab}{a^{2}} \). So, simplify to \( \frac{a(b^{2}+ab)}{a^{2}(b+a)} \), further simplifying gives \( \frac{b^{2}+ab}{a(b+a)} \). b) For \( \frac{3 x^{2}+27}{2 x+6} \times \frac{6 x-18}{x^{4}-81} \), we can factor \( 3(x^{2}+9) \) and \( 2(x+3) \) in the first expression and \( 6(x-3) \) in the second. The denominator \( x^{4}-81 \) factors to \( (x^{2}-9)(x^{2}+9) \) and further to \( (x-3)(x+3)(x^{2}+9) \). Cancel \( (x-3) \) and \( (x+3) \), leading to \( \frac{3 \cdot 6}{2} \), which simplifies to \( 9 \). c) For \( \frac{2 x^{2}-6 x}{x^{2}-1} \div \frac{x-3}{x^{2}-x} \div \frac{1}{3 x+3} \), we factor \( 2x(x-3) \) and \( (x-1)(x+1) \) in the first expression, while the second is \( (x-3)/(x(x-1)) \). The complex division will involve reciprocating the last expression, resulting in some cancellations. After all reductions, you're left with \( \frac{6(x-3)}{(x+1)(3)} \) leading you to the simplified form \( \frac{2(x-3)}{(x+1)} \). d) When simplifying \( \frac{x^{2}+x-6}{3 x^{2}-12 x} \div \frac{x^{3}-2 x^{2}}{x^{2}-16} \times \frac{1}{x+4} \), factor to \( \frac{(x-2)(x+3)}{3x(x-4)} \) for the first and \( \frac{x^{2}(x-2)}{(x-4)(x+4)} \) for the second. Flipping and multiplying gives you a chance to cancel \( (x-2) \) and leads to simplifications that ultimately yield \( \frac{x+3}{3x} \). e) For \( \frac{a^{2}b-ab}{a^{3}+a^{2}} \div \frac{a^{2}b-2ab+b}{a^{3}-a} \), you can factor \( ab(a-b) \) and \( a^{2}(a+1) \) in the first and adjust the second expression to something similar. After simplifications, flipping the second fraction, you're left with a nice product giving you clear insights on how the 'a's and 'b's play along. f) Finally, in \( \frac{a^{2}-2a+1}{a^{2}+2a+1} \times \frac{a^{2}-1}{a^{2}} \times \frac{a^{2}+a}{a^{2}-a} \), factor the first to \( \frac{(a-1)^{2}}{(a+1)^{2}} \), the second to \( \frac{(a-1)(a+1)}{a^2} \) and the third simplifies to \( \frac{a(a+1)}{a(a-1)} \). These lead to valuable cancellations that empower students to grasp fractions better. Enjoy the math journey and keep practicing!