[Math] How to prove $\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$

Question

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?

Answer 1

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.