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?
Proof by contradiction that an expression is irrational
proof-verification
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Best Answer
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.