Question:
Find a nonzero monic polynomial $P(x)$ with integer coefficients and minimal degree such that $P(1-\sqrt[3]2+\sqrt[3]4)=0$.  (A polynomial is called $\textit{monic}$ if its leading coefficient is $1$.)

Answer:
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
\[f(x) = \boxed{x^3 - 3x^2 + 9x - 9}.\]