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.

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.
