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?

Your mistake occurs at this step
$$(3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^3=(81  + 18 \cdot 3^{2/3} \cdot 3^{2/3} + 12 \cdot 3^{1/3} \cdot 3^{4/3} + 72)$$
Here you have forgotten the binomial coefficients $\binom{3}{1}$ and $\binom{3}{2}$.
