multiplicative functions of powers Suppose I have a multiplicative function $f(n),$ and I want to understand the behavior of
$$
\sum_{n<x} f(n^k),
$$ for some integer $k.$ This seems like it should be easy (since the Dirichlet series seems to be just evaluated at $ks$), but it can't be too easy, since, for example, for $f(n) = \mu(n),$ $f(n^k)$ is identically $0$ for every $k>1.$
EDIT I have no idea why this question engendered such a negative reaction.  The perfect answer was given by socalledfriendDon in the comments to the one answer (this is now the accepted answer): under certain conditions, you have an asymptotic formula for average value of multiplicative functions, as given by Wirsing's theorem. Certain conditions are: positivity, having a mean on primes, and having slow growth on prime powers.
For the function $f(n) = \tau(n^k)$ this gives an average value of $\log^k(x).$ 
Many of the pooh-poohers gave completely incorrect references. One gave a reference to Erdos' paper from 1952, where Uncle Paul analyzes the behavior of $\tau(P(n))$ where $P$ is an irreducible polynomial (AND, while Erdos' result is correct, his proof is buggy). A better reference (which no-one gave) is a 1939 paper by van der Corput, but that gives bounds, and not asymptotics.

 A: For each fixed $k$, the function $n\mapsto \tau(n^k)$ is a nonnegative-valued multiplicative function. There are quite general results in the literature about mean values of such functions. One classic paper in this area is
Wirsing, Eduard. Das asymptotische Verhalten von Summen über multiplikative Funktionen. (German) Math. Ann. 143 1961 75–102
His main theorem applies to $\tau(n^k)$ and implies that the partial sums in this case are asymptotic to $c x(\log{x})^k$ for an explicit nonzero constant $c$.
A: Actually it can be arbitrarily hard or arbitrarily easy. 
To elaborate: For a typical multiplicative function $f$ what determines
the behavior of $$
\sum_{n < x} f(n)
$$
is the behavior of $f(p)$ (while the $f(p^2)$ don't have much effect unless the function is extremely large at prime squares). Similarly what determines the behavior of
$$
\sum_{n \leq x} f(n^2)
$$ 
is mostly the behavior of $f(p^2)$. Therefore
if you want the first sum to be easy and the second to be hard take a multiplicative function with say $f(p) = 1$ and $f(p^2) = -1$ (admitedly the Mobius function is hard!). If you want the first sum to  be hard and the second to be easy then take $f(p) = -1$ and $f(p^2) = 1$. Further adjustments can be made if you consider dealing with the Mobius function to be "easy". 
As an example which occurs in real life take $f$ to be the coefficient of a high symmetric power of a Maass form (say the 8-th symmetric power). Then the behavior of $\sum_{n < x} f(n)$ is somewhat understood (but still poorly as we know continuation only up to the 1-line) but that of $\sum_{n < x} f(n^2)$, which (I think) roughly corresponds to a 16-th symmetric power, is a complete mystery...
EDIT: On the other hand if you have something more specific in mind then please give us the details!
