Let $d$ be an integer, that is not a cube of any integer. Show that $d^{1/3}$ is badly approximable iff $d^{2/3}$ is also.
Badly approximable means that, there is a contant $C$ s.t.
$\lvert \beta-\frac pq\rvert<\frac{C}{q^2}$ holds for only finitely many rationals $p/q$ with $p,q$ relatively prime
I already proved that,
$\beta$ is badly approximable iff $n\beta$ and $\frac{1}{\beta}$ are badly aprrox. for $n\in\mathbb N$
(There is a theorem which states that, a real number $\beta$ is badly approximable if the sequence $\{a_i\}_i$, where $a_i$ are the numbers in the continued fraction of $\beta$, i.e. $\beta=[a_0,a_1,\dots]$, is bounded.)
What is the difference between saying that
$d^{1/3}$ badly approx. $\iff$ $d^{2/3}$ is badly approx.
And
$\beta$ badly approx. $\iff$ $\beta^2$ is badly approx.
By what you say you have already proved, $\root3\of d$ badly approximable implies $d^{-1/3}$ badly approximable implies $dd^{-1/3}$ badly approximable, but there's your $d^{2/3}$.