The cube of at least one irrational number is rational
I am supposed to prove the statement above. Here is what I have so far:
Suppose that the cube of at least one irrational number $n$, is rational. By definition of rational, there exists integers $a$ and $b$ such that $r = a/b$ and $b$ is not equal to $0$. Consequently $$\begin{align}n = \sqrt[\Large 3]{\frac ab}\\ n^3 = \frac ab\end{align}$$
I do not know where to go with the proof from here. Is that all I need for the equations and am I doing this right? Or am I going about it completely wrong?