Simplifying $\frac{(\ln a)^3+(\ln b)^3+(\ln c)^3+(\ln d)^3}{\ln ab\;(\ln c\ln d-\ln a\ln b)}$ for positive reals with $abcd=1$ The problem is as follows:

Given the condition for $a$, $b$, $c$, and $d$ to be real positive numbers whose product equals to $1$, find the value of $B$.
$$B=\frac{(\ln a)^3+(\ln b)^3+(\ln c)^3+(\ln d)^3}{\ln ab\cdot\left(\ln c\cdot\ln d-\ln a\cdot\ln b\right)}$$
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{4}\\
2.&\textrm{2}\\
3.&\textrm{3}\\
4.&\textrm{6}\\
\end{array}$

I'm not very sure exactly how to simplify this expression. So far what I've could spot was this:
$abcd=1$
$\ln ab=\ln a + \ln b$
$(\ln a + \ln b)(\ln^2a -\ln a \ln b+\ln^2b)$
$(\ln c + \ln d)(\ln^2c -\ln c \ln d+\ln^2d)$
Taking the natural logarithm to the mentioned expression would become into:
$\ln abcd = \ln 1 = 0$
$\ln a + \ln b + \ln c + \ln d = 0$
But from here is where it becomes quite confusing on how to organize these expressions so that can achieve simplication of the before mentioned equation. Can someone help me here?.
From what I could also spot is:
$(\ln c + \ln d)((\ln c + \ln d)^2 - 3 \ln c \ln d)$
$-(\ln a+\ln b)((\ln a + \ln b)^2 - 3 \ln c \ln d)$
But that's how far I went. Does it exist other manipulation which can be done?

Following the idea given in the comments, let $a=e^A$, $b=e^B$, $c=e^C$, $d=e^D$ then
$$\frac{A^3+B^3+C^3+D^3}{(A+B)\cdot\left(CD-AB\right)}=\frac{(A+B)(A^2-BC+B^2)+(C+D)(C^2-CD+D^2)}{(A+B)\cdot\left(CD-AB\right)}=$$
$$=\frac{(A+B)(A^2-AB+B^2)-(A+B)(C^2-CD+D^2)}{(A+B)\cdot\left(CD-AB\right)}=$$
$$=\frac{(A^2-AB+B^2)-(C^2-CD+D^2)}{\left(CD-AB\right)}=$$
$$=\frac{(CD-AB)+A^2+B^2-C^2-D^2}{\left(CD-AB\right)}=$$
$$=\frac{(CD-AB)+2CD-2AB}{\left(CD-AB\right)}=3$$
using that
$$A+B=-(C+D) \implies \\(A+B)^2=(C+D)^2 \implies A^2+B^2-C^2-D^2=2CD-2AB$$
