How find this value $A+B+C+AB+BC+AC+ABC$ let 
$$A=\dfrac{2\sqrt{3}}{3}i\cdot\cos{\left(\dfrac{\pi}{6}+\dfrac{i}{3}\text{arcsinh}{(\dfrac{3\sqrt{3}}{2})}\right)}$$
$$B=\dfrac{2\sqrt{3}}{3}i\cdot\cos{\left(\dfrac{5\pi}{6}+\dfrac{i}{3}\text{arcsinh}{(\dfrac{3\sqrt{3}}{2})}\right)}$$
$$C=\dfrac{2\sqrt{3}}{3}i\cdot\cos{\left(\dfrac{3\pi}{2}+\dfrac{i}{3}\text{arcsinh}{(\dfrac{3\sqrt{3}}{2})}\right)}$$
Find $$A+B+C+AB+BC+AC+ABC$$
I think we must find $A+B+C=?$
$AB+BC+AC=?$
$ABC=?$
 A: Observe that $$\cos3\left(\frac\pi6+\frac i3\text{arcsinh} \frac{3\sqrt3}2
\right)=\cos\left(\frac\pi2+i\text{arcsinh} \frac{3\sqrt3}2\right)$$
$$=-\sin \left(i\cdot\text{arcsinh} \frac{3\sqrt3}2\right)$$
Now as, $\displaystyle \sin(ix)=i\sinh(x)$
$$-\sin \left(i\cdot\text{arcsinh} \frac{3\sqrt3}2\right)=-i\sinh \left(\text{arcsinh} \frac{3\sqrt3}2\right)=-i \frac{3\sqrt3}2$$
As $\displaystyle\cos3y=4\cos^3y-3\cos y $
So, $\displaystyle\frac A{\dfrac{2\sqrt3}3i}=\cos\left(\frac\pi6+\frac i3\text{arcsinh} \frac{3\sqrt3}2\right)$ is a root of $\displaystyle4t^3-3t=-i \frac{3\sqrt3}2$
If $\displaystyle\cos3P=\cos\left(\frac\pi2+3Q\right),$
$\displaystyle3P=2n\pi\pm \left(\frac\pi2+3Q\right)$ where $n$ is any integer
Taking the '+' sign and $\displaystyle3P=2n\pi+\left(\frac\pi2+3Q\right)=3Q+\frac{(4n+1)\pi}2$
$\displaystyle\implies P=Q+\frac{(4n+1)\pi}6$
Setting $n=0,1,2$ we can see that
$$\displaystyle\frac A{\dfrac{2\sqrt3}3i},\frac B{\dfrac{2\sqrt3}3i}, \frac C{\dfrac{2\sqrt3}3i}$$ are the roots of  $$\displaystyle4t^3-3t=-i \frac{3\sqrt3}2\iff8t^3-6t+i3\sqrt3=0\ \ \ \ (1)$$
Now respond to the invitation of Vieta's formulas 
Method $1:$
$$\displaystyle\frac A{\dfrac{2\sqrt3}3i}+\frac B{\dfrac{2\sqrt3}3i}+ \frac C{\dfrac{2\sqrt3}3i}=0\iff A+B+C=0$$
$$\displaystyle\sum\frac A{\dfrac{2\sqrt3}3i}\cdot\frac B{\dfrac{2\sqrt3}3i}=-\frac68\iff \sum AB=-\frac68\left(\dfrac{2\sqrt3}3i\right)^2=1$$
$$\displaystyle\frac A{\dfrac{2\sqrt3}3i}\cdot\frac B{\dfrac{2\sqrt3}3i}\cdot\frac C{\dfrac{2\sqrt3}3i}=-\frac{i3\sqrt3}8\iff ABC=-\frac{i3\sqrt3}8\left(\dfrac{2\sqrt3}3i\right)^3$$
$$=-i\frac{3\sqrt3}8\frac{8\cdot3\sqrt3i^3}{27}=-i^4=-1$$
Method $2:$
$\displaystyle A+B+C+AB+BC+CA+ABC=-1+\prod(1+A)$
Let $\displaystyle y=1+A\iff \frac{y-1}{\dfrac{2\sqrt3}3i}=\frac A{\dfrac{2\sqrt3}3i}$ which is a root of $(1)$
So, put the value of $\frac A{\dfrac{2\sqrt3}3i}$ in $(1)$ and simplify to form a Cubic equation in $y$
Now, we need $\displaystyle A+B+C+AB+BC+CA+ABC=-1+\prod(1+A)=-1+\prod y$
