3) a) (x²-6x+9)/(x²-3x+9) * (x³+27)/(3x-9)

3) a) $$\frac{x^2-6x+9}{x^2-3x+9} \cdot \frac{x^3+27}{3x-9} = \frac{(x-3)^2}{x^2-3x+9} \cdot \frac{(x+3)(x^2-3x+9)}{3(x-3)} = \frac{(x-3)(x+3)}{3} = \frac{x^2-9}{3}$$

Ответ: (x²-9)/3