Question: Prove that $\displaystyle \sum_{d|n} \tau ^3 (d) = \left( \sum _{d |n} \tau (d) \right)^2$. (Hint: First show that both sides are multiplicative functions.) [$\tau$ is the number of positive divisors of the input.]
Attempt:
LHS: $\tau$ multiplicative $\implies$ $\tau ^3$ multiplicative $\implies$ $\displaystyle \sum_{d|n} \tau ^3 (d)$ multiplicative.
RHS: $\tau$ multiplicative $\implies$ $\displaystyle \sum_{d|n} \tau (d)$ multiplicative $\implies$ $\left( \sum _{d |n} \tau (d) \right)^2$ multiplicative.
This is where I do not know how to proceed. I think it suffices to show that $\displaystyle \sum_{d|p^e} \tau ^3 (d) = \left( \sum _{d |p^e} \tau (d) \right)^2$. Is this the right step? If so, how would I continue this?
EDIT: I have tried continuing the problem with Shark's suggestion. I will post it below:
Note that $\tau (p^k) = k+1$ and $\tau ^3 (p^k) = (k+1)^3$.
It suffices just to prove the identity for $n=p^k$ a prime power.
$$\left(\sum_{d\mid p^k}\tau(d)\right)^2=\left(\sum_{j=0}^k\tau(p^j)\right)^2=\left(\sum_{j=0}^k(j+1)\right)^2$$
$$\sum_{d\mid p^k}\tau(d)^3=\sum_{j=0}^k\tau(p^j)^3=\sum_{j=0}^k(j+1)^3$$.
After some algebra, $\text{LHS} = \frac{1}{4}(k^4 + 6k^3 + 13k^2 + 12k + 4) = \frac{1}{4}(k^4 + 6k^3 + 13k^2 + 12k + 4) = \text{ RHS}$, which finishes the proof.
Some properties involving $\tau :$
$\displaystyle \tau (n) = \sum_{d|n} 1$.
$\tau (p^e) = e+1$.
$\tau$ is multiplicative.
If $f$ is multiplicative then $\displaystyle F(n) = \sum_{d|n} f(d)$.
If $f$ is multiplicative and $n = p_1^{e_1} \cdots p_r^{e_r} $ then $f(n) = f(p_1^{e_1} \cdots p_r^{e_r}) = f(p_1^{e_1})\cdots f(p_r^{e_r})$
As they are multiplicative functions, it suffices just to prove the identity for $n=p^k$ a prime power.
$$\left(\sum_{d\mid p^k}\tau(d)\right)^2=\left(\sum_{j=0}^k\tau(p^j)\right)^2=\left(\sum_{j=0}^k(j+1)\right)^2$$
$$\sum_{d\mid p^k}\tau(d)^3=\sum_{j=0}^k\tau(p^j)^3=\sum_{j=0}^k(j+1)^3$$
I'm sure you can take it from here.