Prove $(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^{-1}$ is an algebraic integer Let $b = (4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^{-1}$, then $1 = b(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})$, and $1 - 4b = b(3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})$.
Therefore $1 - 12b + 48b^2 - 64b^3 = (1 - 4b)^3 = b^3(3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^3$
$= b^3(81  + 18 \cdot 3^{2/3} \cdot 3^{2/3} + 12 \cdot 3^{1/3} \cdot 3^{4/3} + 72) = b^3(153 + 18 \cdot 3 \cdot 3^{1/3} +  18 \cdot 2 \cdot 3^{2/3})$
$= 18b^3(4 +  3 \cdot 3^{1/3} +   2 \cdot 3^{2/3}) + 81b^3 = 18b^3(b^{-1}) + 81b^3= 18b^2 + 81b^3$. Therefore $b$ is a root of $1 - 12b + 48b^2 - 64b^3 - 18b^2 - 81b^3 = - 145b^3 + 30b^2 - 12b + 1$
My problem is $145b^3 - 30b^2 + 12b - 1$ is not monic and is also not reducible, since if it were reducible being of degree $3$ we could factor out a root, which we can't by the rational roots theorem,
Any advice on what I am doing wrong?

I think, there is a mistake in your computations.
We obtain:
$$4b-1+3b\sqrt[3]3+2b\sqrt[3]9=0$$ or
$$(4b-1)^3+81b^3+72b^3-3(4b-1)\cdot18b^2=0$$ or $$b^3+6b^2+12b-1=0.$$
