Solution 1:

I would call this function the $N^{th}$-power free part of $n$, which generalizes the $N=2$ case of the squarefree part of an integer.

Specifically, we have that $$\alpha_N(n)=n\prod_{p^{kN}|n}\left(\frac{1}{p^{N}}\right),$$

and $\alpha_N(n)$ removes all of the $N^{th}$ powers that divide $n$ leaving behind the $N^{th}$-power free part. It will share many properties with the squarefree part. For example, $\alpha_N(n)=n$ for any $N^{th}$-power free integer, which is a set of density $\frac{1}{\zeta(N)}$ in the integers. The average of $\alpha_N(n)$ can be computing using the techniques in the answer Mean Value of a Multiplicative Function close to $n$ in Terms of the Zeta Function. Since $$\mu*\frac{\alpha_{N}(n)}{n}=\begin{cases} 1 & k=0\\ 0 & k\neq0\ \text{mod}\ N\\ -\frac{p^{N}-1}{p^{aN}} & k=aN \end{cases} $$ and we find that $$\sum_{n\leq x}\alpha_N(n)=\frac{x^2}{2}\frac{\zeta(2N)}{\zeta(N)}+O(x\log x).$$ (The error term is $O(x)$ for all $N\geq 3$.) In particular, when $N=2$, we find that the squarefree part of $n$ has average $$\sum_{n\leq x}\alpha_2(n)=\sum_{n\leq x} \text{squarefree}(n)=\frac{x^2\pi^2}{30}+O(x\log x).$$ Similarly, the fourth-powerfree part of $n$ has average $$\sum_{n\leq x}\alpha_4(n)=\sum_{n\leq x}\text{fourth-powerfree-part}(n)=\frac{\pi^4x^2}{210}+O(x).$$