Proof by contradiction that an expression is irrational

Question

The question is: Proove that $\left(\sqrt[3] \frac{q^2-1}{qx}\right)$ is irrational if x is irrational and nonzero and q is a rational number that is not 0 or 1.
I started my proof with: To get a contradiction, suppose that $\left(\sqrt[3] \frac{q^2-1}{qx}\right)$ is rational. Therefore $\left(\sqrt[3] \frac{q^2-1}{qx}\right)$ is in the set of rational numbers. So $\left(\sqrt[3] \frac{q^2-1}{qx}\right)$ = r. I know I now have to proove that x is irrational, but how?
Are these steps correct so far?

Answer 1

From $$r=\frac{c}{d}=\sqrt[3]{\frac{q^2-1}{qx}}.$$ we obtain $$\frac{c^3}{d^3}=\frac{q^2-1}{qx}\Leftrightarrow\frac{c^3q}{d^3(q^2-1)}=\frac{1}{x}.$$ The LHS is rational, since $q,c,d$ are rational. Thus, $1/x$ is rational and $x$ is rational, contradiction. Note that $c^3q$ is an integer and $d^3(q^2-1)$ is a non-zero integer by assumption.

Answer 2

You can continue by solving for $x$ in $\sqrt[3]{\frac{q^2-1}{qx}} = r$ and get a contradiction once you look at what you obtain.

Answer 3

Let $c=\left(\sqrt[3] \frac{q^2-1}{qx}\right)$

Say $c$ is rational meaning we can express $c = \frac mn$, where $m$ and $n$ are integers.

Similarly since $q$ is rational we can write $q = \frac ab$, where $a$ and $b$ are distinct integers with $a \neq 0$.

You can do the algebra (including a cubing step) and express $x$ in terms of the other variables. You should easily be able to reach the required contradiction showing $x$ as a rational expression of two integer expressions.