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.