Proof of $\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$ (not standard proof) I am trying to prove that

$\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$

without using the fact that $\tau$ is multiplicative and products/sums of multiplicative funcions are also multiplicative (here you can find multiple arguments that show this result using such properties).
My approach was the following:

LHS counts the cardinality of the following set:
$$\bigcup_{d| n}\big\{(x,y,z,d)\ /\ x,y,z|d \big\}  $$
and RHS counts this one
$$\big\{(x,d,y,e)\ / \ x|d; \ y|e; \ d,e|n\big\}$$

I tried to establish an explicit bijection between those sets, but I already spent enough time and my biggest efforts on it, unsuccesful. Any idea or hint you can give me?
Thanks in advance! 
 A: Let $n=p_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}$ be a prime factorization of $n$.  Each divisor $d=p_1^{l_1}p_2^{l_2}\cdots p_m^{l_m}$ of $n$ is associated with the vector $\left(l_1,l_2,\ldots,l_m\right) \in \left[k_1\right]\times \left[k_2\right]\times \ldots\times \left[k_m\right]=:S$, where $[t]:=\{0,1,2,\ldots,t\}$ for all integers $t\geq0$.  Partially order the elements of $S$ with the natural order of $\mathbb{Z}$.  In other words, this ordering is given by $\left(l_1,l_2,\ldots,l_m\right)\preceq \left(r_1,r_2,\ldots,r_m\right)$ iff $l_i\leq r_i$ for all $i=1,2,\ldots,m$.  Note that $S$ is a lattice, and we shall write $\vee$ for the join in $S$.  That is, $$\left(l_1,l_2,\ldots,l_m\right)\vee\left(r_1,r_2,\ldots,r_m\right)=\big(\max\left\{l_1,r_1\right\},\max\left\{l_2,r_2\right\},\ldots,\max\left\{l_m,r_m\right\}\big)\,.$$  (If $d$ and $e$ are divisors of $n$ associated to $a$ and $b$ in $S$, respectively, then $\text{lcm}\left(d,e\right)$ is the divisor of $n$ associated to $a\vee b\in S$.) 
Define $$T:=\left\{(a,b)\in S\times S\,\big|\,a\preceq b\right\}\,.$$
Clearly, 
$$|T|=\sum_{b\in S}\,\Big|\big\{a\in S\,\big|\,a\preceq b\big\}\Big|=\sum_{d\mid n}\,\tau(d)\,.$$
We want to count the number of elements of $T\times T$ in two ways.  The first way is the trivial counting:
$$|T\times T|=|T|\cdot |T|=\left(\sum_{d\mid n}\,\tau(d)\right)^2\,.$$
The second way is as follows.  For a fixed $b\in S$, let $N_b$ be the number of $\big((u,v),(x,y)\big)\in T\times T$ such that $v\vee y=b$.  Write $u_j$, $v_j$, $x_j$, $y_j$, and $b_j$ for the $j$-th components of $u$, $v$, $x$, $y$, and $b$, respectively.  For $j=1,2,\ldots,m$, we have $\max\left\{v_j,y_j\right\}=b_j$.  There are $\left(1+v_j\right)\left(1+y_j\right)$ possible pairs $\left(u_j,x_j\right)$ for a given pair $\left(v_j,y_j\right)$.  That is,
$$
\begin{align}
N_b&=\prod_{j=1}^m\,\left(\sum_{v_j=0}^{b_j-1}\,\left(1+v_j\right)\left(1+b_j\right)+\sum_{y_j=0}^{b_j-1}\,\left(1+b_j\right)\left(1+y_j\right)+\left(1+b_j\right)\left(1+b_j\right)\right)
\\
&=\prod_{j=1}^m\,\left(1+b_j\right)\left(2\sum_{r=0}^{b_j-1}\,\left(1+r\right)+\left(1+b_j\right)\right)
\\
&=\prod_{j=1}^m\,\left(1+b_j\right)\Big(b_j\left(1+b_j\right)+\left(1+b_j\right)\Big)=\prod_{j=1}^m\,\left(1+b_j\right)^3=\big(\tau(d)\big)^3\,,
\end{align}$$
if $d$ is the divisor of $n$ associated to $b$.  Therefore,
$$|T\times T|=\sum_{b\in S}\,N_b=\sum_{d\mid n}\,\big(\tau(d)\big)^3\,.$$
A: Edit: the following is wrong, because Yoneda's lemma can't be applied to hom functors restricted to subcategories without some conditions (which aren't attained here). But in the spirit of combinatorial species, I do think category theory is the correct setting in which to explore the question of natural bijections. I have some new back-of-the-envelope ideas I'll look at later.

I submit that there is no natural bijection between the two.
Consider the finite posets $\Lambda=\{x,y,z,d\}$ with $x,y,z\le d$ and $\Gamma=\{x,d\}\cup \{y,e\}$ with $x\le d$ and $y\le e$ and no other relations. These are objects in the category $\sf Pos$ of posets. Our equation states that $\hom_{\sf Pos}(\Lambda,I)$ and $\hom_{\sf Pos}(\Gamma,I)$ are isomorphic in the category $\sf Set$ of sets for any finite interval $I$ of $(\Bbb N,|)$ - meaning the positive naturals ordered by divisibility. Note that every finite interval $I$ of $(\Bbb N,\mid )$ looks like $[a,b]\cong[1,b/a]$.
Now, these two homsets are only equal when we're considering these intervals $I$, so let $\cal L$ be the category of posets isomorphic to an interval $[1,n]$ of $(\Bbb N,\mid)$. Interpret $A=\hom_{\sf Pos}(\Lambda,-)$ and $B=\hom_{\sf Pos}(\Gamma,-)$ as functors $A,B:{\cal L}\to\sf Set$. Is there a natural isomorphism $A\cong B$?
If there were, then $A$ and $B$ would have the same symmetry, i.e. ${\rm Aut}(A)\cong{\rm Aut}(B)$. However according to Yoneda's lemma ${\rm Aut}(A)={\rm Aut}(\Lambda)=S_3$ and ${\rm Aut}(B)={\rm Aut}(\Gamma)=C_2$.
A: Let's have $x \leq d \leq n$ and $ y \leq b \leq n$;  somehow to find $x,y,z \leq d_1 \leq  n $ ... how about $d_1 = \mathrm{lcm}(b,d) $ and $x,y,z \big| d_1$?
Then starting from $x,y,z \leq d_1 \leq n $ what is the inverse? $b = \mathrm{lcm}(x,y,z) $ and $d = d_1/b$.

Added 09-17-15 
Arthur Benjamin - famous for multiplying large numbers in his head - has a few bijective proofs that $(\sum k)^2 = \sum k^3$ 
For the case of perfect prime powers, $27$ has 4 divisors $1,3,9$ and itself.  In general $\tau(p^k) = k+1$. Let:
\begin{array}{cclclcc}
S &=& \{ (a,b,c,d) &:& 0 \leq a,b,c \leq d \leq n & & &\} \\
T &=& \{ \big((x,y),(z,w)\big) &:& 0 \leq x \leq y \leq n & 0 \leq z \leq w \leq n & &\}
\end{array}
The bijection depends on whether $a < b, a = b, a > b$.  We have:
$$ 
(a,b,c,\color{#64C277}{d})\mapsto
\begin{array}{ccc}
\Big( (a,b), (c,\color{#64C277}{d})\Big) & \text{if} & a< b \\
\Big( (b,a), (c,\color{#64C277}{d})\Big) & \text{if} & b<a \\
\Big( (b,d),(c,\color{#64C277}{d})\Big) & \text{if} & a= b \\
\end{array}$$
We can lift this proof to the identity on the divisor function $\tau (d)$:
\begin{array}{cclclcc}
S &=& \{ (a,b,c,d) &:& a,b,c|d & d|n& &\} \\
T &=& \{ \big((x,y),(z,w)\big) &:& x|y, y |n,  & z|w, w |n & &\}
\end{array}
The bijection explits into two parts with $(a,b)$ and $(c,d)$ behaving separately:
$$ (a,b,c,d) \mapsto \Big(\mathrm{gcd}(a,b), \mathrm{lcm}(a,b)\Big), \Big( c,d \Big)$$
