Ring of integers in a cubic extension Let $L=\mathbb{Q}[\alpha]$, with $\alpha^3=10$. How can be proved that
$$\frac{\alpha^2+\alpha+1}{3}$$
is in $O_L$, the ring of integers of $L$?
 A: Factoring $10-1$ in two different ways yields the equality 
$$\frac{\alpha^2+\alpha+1}{3}=\frac{3}{\alpha-1}=\beta.$$
Rescale $(1+3/\beta)^3=10$ so $\beta$ satisfies an integer-coefficient polynomial.
A: It's a root of $x^3-x^2-3x-3$, which is a polynomial with integer coefficients. To see this, use the geometric series formula to sum $\alpha^2+\alpha+1$ and play around with the resulting expression.
A: Method One: Linear Algebra. Set $\beta=(\alpha^2+\alpha+1)/3$. Compute $\beta^2$ and $\beta^3$, writing the results in the form $\square\alpha^2+\square\alpha+\square$. Set $a\beta^3+b\beta^2+c\beta+d=0$, scale by $3$, write as $\square\alpha^2+\square\alpha+\square=0$, set each $\square=0$ for a system of $3$ linear equations in $4$ variables. Use elimination and substitution, or more sophisticated techniques if desired, for a solution, scaling so $a=1$ and $a,b,c,d\in\Bbb Z$.
Method Two: Galois Theory. The cube roots of $10$ are $\alpha,\omega\alpha,\omega^2\alpha$ so these must be its conjugates under Galois action $\sigma$. Write down $\sigma\beta$ and $\sigma^2\beta$ then expand $(x-\beta)(x-\sigma\beta)(x-\sigma^2\beta)=0$.
