Let $$a = \left(\sqrt[3]{2} - 1\right)^{1/3}$$, $$b = \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}$$ then

Let $$t = \sqrt[3]{2}$$, so $$b = \frac{1 - t + t^2}{\sqrt[3]{9}}$$ and $$b^3 = \frac{(1 - t + t^2)^3}{9}$$. Since $$(1 + t)(1 - t + t^2) = 1 + t^3 = 3$$, we get $$b^3 = \frac{27}{9(1 + t)^3} = \frac{3}{(1 + t)^3}$$. Also $$(1 + t)^3 = 1 + 3t + 3t^2 + 2 = 3(1 + t + t^2)$$ and $$(t - 1)(1 + t + t^2) = t^3 - 1 = 1$$, so $$b^3 = \frac{1}{1 + t + t^2} = t - 1 = a^3$$. Hence $$a = b$$ (both are about 0.638).