A formula for $D(x)D(y) - D(xy)$ in terms of the sum-of-aliquot-divisors function, when $\gcd(x,y)=1$ Hereinafter, we shall let $\sigma(z)$ be the sum of divisors of the positive integer $z$.  Denote the deficiency of $z$ by $D(z) = 2z - \sigma(z)$, and the sum of aliquot divisors of $z$ by $s(z) = \sigma(z) - z$.
We shall compute here a formula for $D(x)D(y) - D(xy)$ in terms of the sum-of-aliquot-divisors function, when $\gcd(x,y)=1$.
Suppose that $\gcd(x,y)=1$.
Then we have
$$D(x)D(y) - D(xy) = (2x - \sigma(x))(2y - \sigma(y)) - (2xy - \sigma(xy))$$
$$= 4xy - 2y\sigma(x) - 2x\sigma(y) + \sigma(x)\sigma(y) - 2xy + \sigma(x)\sigma(y),$$
where we have used the condition $\gcd(x,y)=1$ in the last equation to derive $\sigma(xy)=\sigma(x)\sigma(y)$.
This gives
$$D(x)D(y) - D(xy) = 2xy - 2y\sigma(x) - 2x\sigma(y) + 2\sigma(x)\sigma(y)$$
so that we obtain
$$D(x)D(y) - D(xy) = 2y\bigg(x - \sigma(x)\bigg) - 2\sigma(y)\bigg(x - \sigma(x)\bigg)$$
which simplifies to
$$D(x)D(y) - D(xy) = 2\bigg(x - \sigma(x)\bigg)\bigg(y - \sigma(y)\bigg) = 2\bigg(\sigma(x) - x\bigg)\bigg(\sigma(y) - y\bigg) = 2s(x)s(y).$$
Here are my inquiries:
QUESTIONS

(1) Is it possible to extend the formula
$$D(x)D(y) - D(xy) = 2s(x)s(y)$$
to, say, something that uses three or more arguments (which are pairwise coprime)?


(2) If the answer to Question (1) is YES, what is the closed form for the formula and how can it be proved, in general?

 A: Let $s(x_i):=s_i$.
If $\gcd(x_i,x_j)=1$ for any $(i,j)$ where $1\le i\lt j\le n$, then we have
$$\begin{align}\prod_{i=1}^{n}D(x_i)-D\bigg(\prod_{i=1}^{n}x_i\bigg)
&=\prod_{i=1}^{n}(2x_i-\sigma(x_i))-2\prod_{i=1}^{n}x_i+\sigma\bigg(\prod_{i=1}^{n}x_i\bigg)
\\\\&=\prod_{i=1}^{n}(2x_i-\sigma(x_i))-2\prod_{i=1}^{n}x_i+\prod_{i=1}^{n}\sigma(x_i)
\\\\&=\prod_{i=1}^{n}(x_i-s_i)+\prod_{i=1}^{n}(x_i+s_i)-2\prod_{i=1}^{n}x_i
\\\\&=\begin{cases}\displaystyle 2\bigg(\sum_{(n-2,2)}+\sum_{(n-4,4)}+\cdots +\sum_{(0,n)}\bigg)&\text{if $n$ is even}\\\\\displaystyle 2\bigg(\sum_{(n-2,2)}+\sum_{(n-4,4)}+\cdots +\sum_{(1,n-1)}\bigg)&\text{if $n$ is odd}\end{cases}\end{align}$$
where $\displaystyle\sum_{(i,j)}$ represents the sum of all the terms which are of the form $$\underbrace{xx\cdots xx}_{i}\underbrace{ss\cdots ss}_{j}$$
with $i+j=n$.

For example, for $n=4$, we have
$$\prod_{i=1}^{4}D(x_i)-D\bigg(\prod_{i=1}^{4}x_i\bigg)$$
$$\small=2\bigg(x_1x_2s_3s_4+x_1x_3s_2s_4+x_1x_4s_2s_3+x_2x_3s_1s_4+x_2x_4s_1s_3+x_3x_4s_1s_2+s_1s_2s_3s_4\bigg)$$
For $n=5$, we have
$$\prod_{i=1}^{5}D(x_i)-D\bigg(\prod_{i=1}^{5}x_i\bigg)$$
$$=2\bigg(x_3x_4x_5s_1s_2+x_2x_4x_5s_1s_3+x_2x_3x_5s_1s_4+x_2x_3x_4s_1s_5+x_1x_4x_5s_2s_3+x_1x_3x_5s_2s_4+x_1x_3x_4s_2s_5+x_1x_2x_5s_3s_4+x_1x_2x_4s_3s_5+x_1x_2x_3s_4s_5+x_1s_2s_3s_4s_5+x_2s_1s_3s_4s_5+x_3s_1s_2s_4s_5+x_4s_1s_2s_3s_5+x_5s_1s_2s_3s_4\bigg)$$
A: Here is my own attempt for the case of three ($3$) arguments.
Suppose that
$$\gcd(x,y)=\gcd(x,z)=\gcd(y,z)=1.$$
Then we have
$$D(x)D(y)D(z) - D(xyz) = (2x-\sigma(x))(2y-\sigma(y))(2z-\sigma(z))-(2xyz-\sigma(xyz))$$
$$=(4xy-2y\sigma(x)-2x\sigma(y)+\sigma(x)\sigma(y))(2z-\sigma(z))-2xyz+\sigma(x)\sigma(y)\sigma(z)$$
$$=8xyz-4yz\sigma(x)-4xz\sigma(y)+2z\sigma(x)\sigma(y)-4xy\sigma(z)+2y\sigma(x)\sigma(z)+2x\sigma(y)\sigma(z)-\sigma(x)\sigma(y)\sigma(z)-2xyz+\sigma(x)\sigma(y)\sigma(z)$$
$$=2xyz-2yz\sigma(x)-2yz\sigma(x)+2z\sigma(x)\sigma(y)$$
$$+2xyz-2xz\sigma(y)-2xz\sigma(y)+2x\sigma(y)\sigma(z)$$
$$+2xyz-2xy\sigma(z)-2xy\sigma(z)+2y\sigma(x)\sigma(z),$$
from which we obtain
$$=2yz(x-\sigma(x))-2z\sigma(x)(y-\sigma(y))$$
$$+2xz(y-\sigma(y))-2x\sigma(y)(z-\sigma(z))$$
$$+2xy(z-\sigma(z))-2y\sigma(z)(x-\sigma(x))$$
from which we get
$$=2y(x-\sigma(x))(z-\sigma(z))+2z(y-\sigma(y))(x-\sigma(x))+2x(z-\sigma(z))(y-\sigma(y)).$$

This finally gives the formula
$$D(x)D(y)D(z)-D(xyz)=2\bigg(xs(y)s(z)+ys(x)s(z)+zs(x)s(y)\bigg).$$


Checking the formula for $(x,y,z)=(3,5,7)$ gives
$$D(x)D(y)D(z)-D(xyz)=D(3)D(5)D(7)-D(105)=2\cdot{4}\cdot{6}-18=48-18=30$$
$$2\bigg(xs(y)s(z)+ys(x)s(z)+zs(x)s(y)\bigg)=2\bigg(3\cdot s(5)s(7)+5\cdot s(3)s(7)+7\cdot s(3)s(5)\bigg)=2\bigg(3\cdot{1}\cdot{1}+5\cdot{1}\cdot{1}+7\cdot{1}\cdot{1}\bigg)=2\cdot{15}=30.$$

