3. Let's find all mutually prime triples $x, y, (x+y)$ of divisors of the number $a=1944=2^{3} \cdot 3^{5}$. Note that if $p>1$ is a divisor of $x$ and $y$, then $p$ divides $x+y$. Similarly, if $p>1$ is a divisor of $x$ and $x+y$, then it divides $y$, and any divisor $p>1$ of the pair $y$ and $x+y$ divides $x$. Thus, the numbers $x, y, (x+y)$ have different prime divisors, except for 1. Since there are only two such divisors, namely 2 and 3, and there are three numbers, one of them must be 1 ( $x+y \neq 1$ ).

Case 1. $x=1, y=2 m, x+y=y+1=3 n, n, m \in Z, m \geq 1, n \geq 1$

Then $3 n-2 m=1 \rightarrow\left\{\begin{array}{l}n=1+2 s \\ m=1+3 s\end{array}, s \in Z \rightarrow\left\{\begin{array}{c}y=6 s+2 \\ x+y=6 s+3\end{array}\right.\right.$. In this case, $y$ can only contain powers of two, i.e., $y=2,4,8$, and $x+y$ - powers of three, i.e., $x+y=3,9,27,81$. The latter is possible for $s=0$ and $s=1$.
The first solution $x=1, y=2, x+y=3$. This coprime triplet of divisors of the number $a=1944=2^{3} \cdot 3^{5}$ generates triplets $x=t, y=2 t, x+y=3 t$, where $t$ is any divisor of the number, where $b=324=2^{2} 3^{4}$. The largest such divisor is $t=324$ and $x \cdot y \cdot(x+y)_{\max }=6 \cdot 324^{3}=2^{7} \cdot 3^{13}$.

The second solution $x=1, y=8, x+y=9$ will generate triplets $x=t, y=8 t, x+y=9 t$, where $t-$ is any divisor of the number, where $c=27=3^{3}$. The largest divisor is $t=27$, so $x \cdot y \cdot(x+y)_{\max }=72 \cdot 27^{3}=2^{3} \cdot 3^{11}$

Case 2. $x=1, y=3 m, x+y=y+1=2 n, m, n \in Z, m \geq 1, n \geq 1$

$2 n-3 m=1 \rightarrow\left\{\begin{array}{l}n=3 s+2 \\ m=2 s+1\end{array} \rightarrow\left\{\begin{array}{c}y=6 s+3 \\ x+y=6 s+4\end{array}\right.\right.$. Possible values for $y=3,9,27,81$,

for $x+y=y+1=2,4,8$. There is only one solution corresponding to the value $s=0 \rightarrow x=1, y=3, x+y=4$. This coprime triplet generates valid triplets $x=t, y=3 t, x+y=4 t$. where $t-$ is any divisor of the number $d=162=2 \cdot 3^{4}$. The maximum divisor $t=162$ corresponds to the maximum product $x \cdot y \cdot(x+y)_{\max }=12 \cdot 162^{3}=2^{5} \cdot 3^{13}$

Case 3. $x=1, y=1, x+y=2$

This coprime triplet corresponds to valid triplets $x=t, y=t, x+y=2 t$, where $t-$ is any divisor of the number $e=972=2^{2} \cdot 3^{5}$. The maximum divisor $t=972$ corresponds to the maximum product $x \cdot y \cdot(x+y)_{\max }=2 \cdot 972^{3}=2^{7} \cdot 3^{15}$.

The numbers $x$ and $y$ enter the condition symmetrically, so cases with $y=1$ do not create new products of triplets $x y(x+y)$.

Answer: $x \cdot y \cdot(x+y)_{\max }=2 \cdot 972^{3}=2^{7} \cdot 3^{15}$