Let $ x = 1 - \sqrt[3]{2} + \sqrt[3]{4} $. Note that $ (1 - \sqrt[3]{2} + \sqrt[3]{4})(1 + \sqrt[3]{2}) = 3, $ so \[x = \frac{3}{1 + \sqrt[3]{2}}.\]Then \[\frac{3}{x} = 1 + \sqrt[3]{2},\]so \[\frac{3}{x} - 1 = \frac{3 - x}{x} = \sqrt[3]{2}.\]Cubing both sides, we get \[\frac{-x^3 + 9x^2 - 27x + 27}{x^3} = 2,\]so $ -x^3 + 9x^2 - 27x + 27 = 2x^3 $. This simplifies to $ 3x^3 - 9x^2 + 27x - 27 = 3(x^3 - 3x^2 + 9x - 9) = 0, $ so we can take \[f(x) = \boxed{x^3 - 3x^2 + 9x - 9}.\]