How to prove that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer As the title says, I'm trying to show that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer.  
I suppose there's probably some heavy duty classification theorems that give one line proofs to this but I don't have any of that at my disposable so basically I'm trying to construct a polynomial over $\mathbb{Z}$ which has this complex number as a root.
My general strategy is to raise both sides of the equation 
$$x = \frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$$
to the $n^{th}$ power and then break up the resulting sum in such a way as to resubstitute back in smaller powers of $x$.  Also since this root is complex I know it must come in a conjugate pair for the coefficients of my polynomial to be real, thus I know that
$$x = -\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$$
must also be a root of my polynomial.  Hence from this I obtain:
$$3x^2 = (10^{\frac{2}{3}} -1 )^2$$
However since my root is pure imaginary I don't really get any more information from this, so I'm a bit stumped, I tried raising both sides of $3x^2 -1 = 10^{\frac{4}{3}} - (2)10^{\frac{2}{3}}$ to the third power but it doesn't look like it's going to break up correctly, can anyone help me with this?  Thanks.
 A: Your general strategy is workable, but you'll be a lot happier if you start by eliminating the cube root. Start with
$$x \sqrt{-3}+1=10^{2/3} \, .$$
Cube this equation, expand the left-hand side, collect all the factors of $\sqrt{-3}$ on one side, and square the result, and you'll be left with a sextic (hopefully the same one Robert came up with, up to a constant multiple).
A: If we start with $\sqrt{-3}x + 1 = 10^{2/3}$, we get
$$(\sqrt{-3}x+1)^3 = 100$$
hence
$$100  = -3\sqrt{-3}x^3 - 9x^2 + 3\sqrt{-3}x + 1.$$
Therefore,
$$3\sqrt{-3}x^3 + 9x^2 - 3\sqrt{-3}x + 99 = 0.$$
Dividing through by $3\sqrt{-3}$ we obtain
$$x^3 + \frac{3}{\sqrt{-3}}x^2 - x + \frac{33}{\sqrt{-3}}=0$$ 
and rationalizing we get
$$x^3 + \frac{3\sqrt{-3}}{-3}x^2 - x + \frac{33\sqrt{-3}}{-3} = 0$$
or
$$x^3 - \sqrt{-3}x^2 - x -11\sqrt{-3}=0.$$
This is a monic polynomial with coefficients in $\mathbb{Z}[\sqrt{-3}]$; hence $x$ is integral over $\mathbb{Z}[\sqrt{-3}]$, which in turn is integral over $\mathbb{Z}$, so $x$ is integral over $\mathbb{Z}$, as desired.
To go from this to the polynomial in Robert Israel's answer, we multiply by the conjugate:
$$\begin{align*}
&\Bigl((x^3-x) - \sqrt{-3}(x^2+11)\Bigr)\Bigl((x^3-x)+\sqrt{-3}(x^2+11)\Bigr)\\
&\qquad = (x^3-x)^2 + 3(x^2+11)^2\\
&\qquad = x^6 - 2x^4 + x^2 + 3x^4 + 66x^2 + 363\\
&\qquad = x^6 + x^4 + 67x^2 + 363, 
\end{align*}$$
and we are done.
A: A somewhat less computational approach can be based on the idea that an algebraic number is an algebraic integer if and only if it is everywhere locally integral. Certainly $\sqrt{-3}$ is a $p$-adic unit (possibly in a quadratic extension) everywhere except $3$, so the only issue is at $3$.
Second, observe/speculate that there is a cube root $\alpha$ of $10$ in $\mathbb Q_3$, using Hensel's Lemma: $10$ is pretty close to $1$ modulo powers of $3$, that is, $3$-adically. A useful form of Hensel's lemma here is that a monic polynomial $f$ with coefficients in $\mathbb Z_3$ and $x_o\in \mathbb Z_3$ such that $|f(x_o)/f'(x_o)^2|_3<1$ produces a root of $f(x)=0$ in $\mathbb Z_3$.
From this, we see that if we can find a cube root of $10$ mod $3^3=27$, we have a $3$-adic cube root. Indeed, $4^3=64=10+2\cdot 3^3$.
So there is $\alpha\in \mathbb Z_3$ with $\alpha-4=0 \mod 3$, so ${\alpha-1\over 3}$ is a $3$-adic integer, and certainly ${\alpha-1\over \sqrt{-3}}$ is integral over $\mathbb Z_3$, although in a quadratic extension.
A similar pattern occurs for all odd primes: $(1+p)^p=1+p^2 \mod p^3$, so there is a $p$-th root of $1+p^2$ in $\mathbb Z_p$.
A: A systematic (but possibly not particularly clever) approach would be to work in $\mathbb Q[\alpha,\beta]$ where $\alpha=\sqrt[3]{10}$ and $\beta=i\sqrt{3}$.
First note that $1/\beta = \frac{-1}{3}\beta$. Then your $x$ is some polynomial in $\alpha$ and $\beta$. Write out the powers of $x$ up to $x^6$ as similar polynomials, simplifying along the way using $\alpha^3=10$ and $\beta^2=-3$.
You now have $1, x, x^2,\ldots x^6$ expressed as polynomials of degree at most $3$ in $\alpha$ and at most $2$ in $\beta$. The space of such polynomials is a $6$-dimensional rational vector space, so you can find a nontrivial linear combination of your $7$ polynomials -- with is then a polynomial in $x$ that has your mystery number as a root.
A: $$\begin{eqnarray}  & &\rm x &=&\,\rm \frac{\sqrt[3]{a}-1}{\sqrt{b}} \\
&\Rightarrow& 0 &=&\,\rm (\sqrt{b}\,x + 1)^3\!-a\, =\, 3b\,x^2\!+1\!-\!a + (bx^3\! + 3x)\, \color{#C00}{\sqrt{b}}\ =:\ f(\color{#C00}{\sqrt{b}}) \\
 &\Rightarrow& 0 & = &\,\rm f(\color{#C00}{\sqrt{b}})\,f(\color{#C00}{-\sqrt{b}})\, =\, -b^3\,x^6 + 3b^2\,x^4 -3(2a\!+\!1)b\,x^2+(a\!-\!1)^2\\
\rm a=100,\ b = -3 &\Rightarrow& 0 &=&\,\rm 27\,(x^6 + x^4 + 67\, x^3 + 363) \\
\rm a=10,\ \ \ b = -3 &\Rightarrow\,&  0 &=&\,\rm 27\, (x^6+x^4 +\,\ 7\, x^2 + 3)\\
\end{eqnarray}
$$
A: The minimal polynomial, found with help from Maple, is $z^6+z^4+67 z^2+363$.
