11) 1/9 a² - 2/9 ab + 1/9 b² =

We can factor out $$\frac{1}{9}$$ from the expression: $$\frac{1}{9}a^2 - \frac{2}{9}ab + \frac{1}{9}b^2 = \frac{1}{9}(a^2 - 2ab + b^2)$$ Now, the expression inside the parentheses is a perfect square trinomial: $$a^2 - 2ab + b^2 = (a-b)^2$$ Therefore, $$\frac{1}{9}a^2 - \frac{2}{9}ab + \frac{1}{9}b^2 = \frac{1}{9}(a-b)^2 = \frac{(a-b)^2}{9} = (\frac{a-b}{3})^2$$ Answer: $$\frac{1}{9}(a - b)^2$$ or $$(\frac{a-b}{3})^2$$