66. 3) $$\frac{c^2-8}{2c+3} - \frac{16c-2c^3}{9-4c^2};$$

$$\frac{c^2-8}{2c+3} - \frac{16c-2c^3}{9-4c^2} = \frac{c^2-8}{2c+3} + \frac{16c-2c^3}{4c^2-9} = \frac{c^2-8}{2c+3} + \frac{16c-2c^3}{(2c-3)(2c+3)} = \frac{(c^2-8)(2c-3)}{(2c+3)(2c-3)} + \frac{16c-2c^3}{(2c-3)(2c+3)} = \frac{2c^3-3c^2-16c+24+16c-2c^3}{(2c-3)(2c+3)} = \frac{-3c^2+24}{(2c-3)(2c+3)} = \frac{-3(c^2-8)}{4c^2-9}$$ Ответ: $$\frac{-3(c^2-8)}{4c^2-9}$$