For integers $a,b\geq 1$ we denote with $(a,b)$ their greatest common divisor, and the $\alpha$th root with $ \sqrt[\alpha]{x} =x^{1/\alpha}$, where $\alpha\geq 2$. Here with $\mu(n)$ we denote the Möbius function. Then I did calculations following the ideas of [1], for a different example, to get
$$\sum_{\substack{n\leq x\\(n,k)=1}}\frac{1}{\sqrt[\alpha]{n}}=\frac{1}{\alpha} \left( \sum_{d\mid k} \frac{\mu(d)}{\sqrt[\alpha]{d}} \right)\log x+O(1),$$ where $k$ denotes a positive number. But I don't know how do calculations for $$\sum_{d\mid k} \frac{\mu(d)}{\sqrt[\alpha]{d}}$$ with the purpose to obtain a more eleaborated and concise result.
Question For $\alpha\geq 2$, what's your approach to get the asymptotic behaviour of $$\sum_{\substack{n\leq x\\(n,k)=1}}\frac{1}{\sqrt[\alpha]{n}},$$ as $x$ is large?. Many thanks.
My thoughts were about Möbius inversion, but I didn't get nothing useful.
[1] Murty, Problems in Analytic Number Theory, Second Edition, GTM 206 Springer (2008). I am saying that the ideas of my calculations were from exercises likes to Exercise 1.5.1 or Exercise 1.5.8.