Simplify the following fractions (assume all denominators are non-zero): \( \begin{array}{ll}\text { a) } \frac{x^{2}-y^{2}}{3} \times \frac{6 x-3 y}{x^{3}+y^{3}} & \text { b) } \frac{2 x^{3}-16 y^{3}}{4 x+8 y} \times \frac{x^{2}-4 y^{2}}{x^{2}+2 x y+4 y^{2}} \\ \text { c) } \frac{x^{2}+5 x+6}{2 x+4} \times \frac{x-3}{x^{2}-9} & \text { d) } \frac{x^{2}+3 x-4}{x^{3}+64 y^{3}} \div \frac{x^{2}-1}{x^{2}-4 x y+16 y^{2}} \\ \text { e) } \frac{x^{2}+4 x-45}{x^{2}-25} \div \frac{x^{2}-5 x+25}{x^{3}+125} & \text { f) } \frac{a x+a y}{x^{2}-y^{2}} \times \frac{x^{2}+2 x y-3 y^{2}}{a} \\ \text { g) } \frac{x^{2}-16}{x^{3}+64} \div \frac{3 x-12}{x^{2}-4 x+16} & \text { h) } \frac{x^{2}-6 x+5}{x^{2}+x+1} \div \frac{3 x-15}{x^{3}-1} \\ \text { i) } \frac{8 x^{3}-27}{8 x-12} & \text { f) } \frac{x^{2}-5 x+6}{x^{3}-9} \div \frac{x^{2}-9}{x^{3}+2 x^{2}+4 x}\end{array} \)

Let's simplify the fractions step-by-step: a) \[ \frac{x^{2}-y^{2}}{3} \times \frac{6x-3y}{x^{3}+y^{3}} = \frac{(x-y)(x+y)}{3} \times \frac{3(2x-y)}{(x+y)(x^2 - xy + y^2)} \] We can cancel \( x+y \) from the numerator and the denominator: \[ = \frac{(x-y)(2x-y)}{1} = (x-y)(2x-y) \] b) \[ \frac{2x^{3}-16y^{3}}{4x+8y} \times \frac{x^{2}-4y^{2}}{x^{2}+2xy+4y^{2}} = \frac{2(x^{3}-8y^{3})}{4(x+2y)} \times \frac{(x-2y)(x+2y)}{(x+y)^2} \] Now we can factor and simplify: \[ = \frac{(x-2y)(x^2+2xy+4y^2)(x-2y)(x+2y)}{2(x+2y)(x+y)^2} \] Cancel \( x+2y \) and we finally get: \[ = \frac{(x-2y)(x-2y)(x+2y)}{2(x+y)^2} = \frac{(x-2y)^2(x+2y)}{2(x+y)^2} \] c) \[ \frac{x^{2}+5x+6}{2x+4} \times \frac{x-3}{x^{2}-9} = \frac{(x+2)(x+3)}{2(x+2)} \times \frac{x-3}{(x-3)(x+3)} \] Cancel \( x+2 \) and \( x-3 \): \[ = \frac{1}{2} = \frac{1}{2} \] d) \[ \frac{x^{2}+3x-4}{x^{3}+64y^{3}} \div \frac{x^{2}-1}{x^{2}-4xy+16y^{2}} = \frac{(x+4)(x-1)}{(x+4)(x^2-4y^2)} \times \frac{x^2-4xy+16y^2}{(x-1)(x^2-1)} \] Here we can cancel \( x+4 \) and \( x-1 \): \[ = \frac{x^2-4xy+16y^2}{x^2-1} = (x-1)(x^2-1) \] e) \[ \frac{x^{2}+4x-45}{x^{2}-25} \div \frac{x^{2}-5x+25}{x^{3}+125} = \frac{(x+9)(x-5)}{(x-5)(x+5)} \times \frac{x^{3}+125}{x^{2}-5x+25} \] Now cancel \( x-5 \): \[ = \frac{x+9}{x+5} \times \text{(rest of the terms)} = \frac{(x+9)(x^3+125)}{(x+5)(x^2-5x+25)} \] f) \[ \frac{ax+ay}{x^{2}-y^{2}} \times \frac{x^{2}+2xy-3y^{2}}{a} = \frac{a(x+y)}{(x-y)(x+y)} \times \frac{(x-y)(x+3y)}{a} \] We can cancel \( a \) and \( x+y \): \[ = \frac{x+3y}{x-y} \] g) \[ \frac{x^{2}-16}{x^{3}+64} \div \frac{3x-12}{x^{2}-4x+16} = \frac{(x-4)(x+4)}{(x+4)(x^2+4xy+16y^2)} \times \frac{x^2-4x+16}{3(x-4)} \] Cancelling terms we get: \[ = \frac{x^{2}-4x+16}{3(x+4)} \] h) \[ \frac{x^{2}-6x+5}{x^{2}+x+1