How to prove $\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$ 
For every positive integer $d$, we let $\tau\left(d\right)$ be the number of positive divisors of $d$.
Prove that
\begin{align}
\sum_{d|n} \tau^3(d)
= \left(\sum_{d|n} \tau (d)\right)^2
\end{align}
for each positive integer $n$, where the sums range over all positive divisors $d$ of $n$.

Now I only know that both sides are multiplicative arithmetic functions in $n$. Could you tell me what I need to do next?
 A: If you've shown that both the LHS and the RHS are multiplicative functions, then you must now show it's true for arbitrary prime powers $n=p^r$. In doing so, use $\tau(n)=\sigma_1(p^r)=r+1$ and this. It then follows for all composite numbers by prime factorizing both sides through the multiplication.
A: Recall that if $f$ is a multiplicative function, and
$$g(n)=\sum_{d|n}f\left(\frac{n}{d}\right),$$
then $g$ is a multiplicative function.
From this you can deduce that both sides of your identity are multiplicative functions. Thus to verify the identity, all you need to do is to verify it for $n=p^k$, where $p$ is prime.
For $n=p^k$, the left-hand side is $1^3+2^3+\cdots +k^3+(k+1)^3$ (since $p^i$ has $i+1$ positive divisors for each $i \geq 0$).  The right-hand side is $(1+2+\cdots+k+(k+1))^2$.  The fact that the two are equal is a probably familiar fact. If it isn't, it can be proved by induction.
A: To get this result you need to show 
(a) it is true for prime powers, which you can show with $$ \sum_1^n i^3 = \frac{n^2(n+1)^2}{4}  = \left(\sum_1^n i\right)^2$$
(b) both the left and right sides are multiplicative, which you say you know 
