Does $x^3 - \frac{m}{n}\sqrt{5}x - 1$ has rational root?

Question

I am trying to show whether $p(x) = x^3 - \frac{m}{n}\sqrt{5}x - 1$ has a rational root or not, where $\frac{m}{n}$ is rational. My attempt so far is to turn $p(x)$ into another polynomial $q(x) = ( - \frac{m}{n}\sqrt{5}x - (1 - x^3))(- \frac{m}{n}\sqrt{5}x + (1 - x^3))= -n^2x^6 + 2n^2x^3 + 5m^2x^2 - n^2$, that has all the roots of $p(x)$, and trying to show $p(x)$ has (or has no) rational roots. But I am having trouble determining this for $q(x)$ since the coefficients of the highest and lowest ordered term is an arbitrary integer, which makes applying rational root theorem difficult. So I am wondering whether there is a way to answer the question.

Answer 1

Let $x$ be a rational root.

Thus, $x\neq0$ and $$\sqrt5=\frac{x^3-1}{\frac{m}{n}x},$$ which is a contradiction for $\frac{m}{n}\neq0$.

Can you end it now?