How to write $\zeta(s)^3$ as a function of the generalized Mobius function? Popovici (1963) (see link), created a way to extend the Mobius function, $\mu(n)$, to the complex plane.
The Mobius $\mu(n)$ function is such that:
$\frac{1}{\zeta{(s)}}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}$
With Popovici's extension, $\mu_k(n)$, we get a generalization:
$\frac{1}{\zeta{(s)^k}}=\sum_{n=1}^{\infty}\frac{\mu_k(n)}{n^s}$, where $μ_k=μ∗...∗μ$ taken $k$ times on a Dirichlet convolution is Popovici’s function.
My question is..., how about $\zeta{(s)}^3$? Does his generalization of $\mu(n)$ work in this case? 
And what the function $\mu_{-3}(n)$ would be in this case?
Popovici's generalized Mobius function
 A: Well, let's just look at $\zeta(s)^2$ first:
$$\zeta(s)^2 = \left(1+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\cdots\right)^2$$
So, how many times would we get the term $\tfrac1{n^2}$? Well, exactly as much as it has divisors; we would get $\tfrac1{6^s}$ exactly $4$ times because $6$ has $4$ divisors. So, we have the terms $\tfrac1{6^2}\tfrac1{1^s}$, $\tfrac1{3^2}\tfrac1{2^s}$, etcetera. 
So, if $\tau(n)$ denotes the number of divisors, then $\zeta(s)^2=\sum \frac{\tau(n)}{n^2}$. Does the generalized Möbius function satisfy $\mu_{-2}(n) = \tau(n)$? From the definition wikipedia gives,

$$\mu_k(p^a) = (-1)^a{k\choose a}$$

That doesn't really make any sense for $k < 0$, so I'll just say no, that generalization doesn't work for negative $n$. However, feel free to extend it even more; Popovici's generalized Möbius function isn't that well known, so you could just define your own to mean what you need.

To adress what $\zeta(s)^3$ is; the function you're looking for would be $\tau_3(n)$: more info on oeis.org/A007425.
A: Popovici's generalised Möbius function $\mu_k$ is multiplicative, so is defined by its values on $p^a$ with $p\in\Bbb P,\,a\in\Bbb Z_{\ge 0}$ viz. $$\mu_k(p^a)=(-1)^a\binom{k}{a}=\frac{(-1)^a\prod_{j=0}^{a-1}(k-j)}{a!}.$$The last expression can be evaluated for $k=-3$, viz.$$\mu_{-3}(p^a)=\frac{(a+2)!}{a!2}=\frac{(a+1)(a+2)}{2}.$$Indeed, the binomial theorem implies $$(1-p^{-s})^{-3}=\sum_{a=0}^\infty\frac{(a+1)(a+2)}{2}p^{-as},$$to which we can apply $\prod_{p\in\Bbb P}$ to prove $(\zeta(s))^3=\sum_{n\ge 1}\mu_{-3}(n)n^{-s}$.
