Question:
Find a monic cubic polynomial $P(x)$ with integer coefficients such that
\[P(\sqrt[3]{2} + 1) = 0.\](A polynomial is monic if its leading coefficient is 1.)

Answer:
Let $x = \sqrt[3]{2} + 1.$  Then $x - 1 = \sqrt[3]{2},$ so
\[(x - 1)^3 = 2.\]This simplifies to $x^3 - 3x^2 + 3x - 3 = 0.$  Thus, we can take $P(x) = \boxed{x^3 - 3x^2 + 3x - 3}.$