The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$
$\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1$
Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation $6$ units right, so they become the solutions to the equation $z^8=81.$
We rewrite $z$ to the polar form \[z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,\] where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$
By De Moivre's Theorem, we have \[z^8=r^8\operatorname{cis}(8\theta)={\sqrt3}^8(1),\] from which


$r^8={\sqrt3}^8,$ so $r=\sqrt3.$
$\begin{cases} \begin{aligned} \cos(8\theta) &= 1 \\ \sin(8\theta) &= 0 \end{aligned}, \end{cases}$ so $\theta=0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{5\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}.$

In the complex plane, the solutions to the equation $z^8=81$ are the vertices of a regular octagon with center $0$ and radius $\sqrt3.$
The least possible area of $\triangle ABC$ occurs when $A,B,$ and $C$ are the consecutive vertices of the octagon. For simplicity purposes, let $A=\sqrt3\operatorname{cis}\frac{\pi}{4}=\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i, B=\sqrt3\operatorname{cis}\frac{\pi}{2}=\sqrt3i,$ and $C=\sqrt3\operatorname{cis}\frac{3\pi}{4}=-\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i,$ as shown below.
[asy] /* Made by MRENTHUSIASM */ size(200);   int xMin = -2; int xMax = 2; int yMin = -2; int yMax = 2; int numRays = 24;  //Draws a polar grid that goes out to a number of circles  //equal to big, with numRays specifying the number of rays:  void polarGrid(int big, int numRays)  {   for (int i = 1; i < big+1; ++i)   {     draw(Circle((0,0),i), gray+linewidth(0.4));   }   for (int i=0;i<numRays;++i)    draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); }  //Draws the horizontal gridlines void horizontalLines() {   for (int i = yMin+1; i < yMax; ++i)   {     draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));   } }  //Draws the vertical gridlines void verticalLines() {   for (int i = xMin+1; i < xMax; ++i)   {     draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));   } }  horizontalLines(); verticalLines(); polarGrid(xMax,numRays); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("Re",(xMax,0),2*E); label("Im",(0,yMax),2*N);  //The n such that we're taking the nth roots of unity multiplied by 2. int n = 8;  pair A[]; for(int i = 0; i <= n-1; i+=1) {   A[i] = rotate(360*i/n)*(sqrt(3),0); }  label("$A$",A[1],1.5*NE,UnFill); label("$B$",A[2],1.5*NE,UnFill); label("$C$",A[3],1.5*NW,UnFill); fill(A[1]--A[2]--A[3]--cycle,green);  draw(A[1]--A[3]^^A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,red);  for(int i = 0; i< n; ++i) dot(A[i],red+linewidth(4.5));  [/asy]
Note that $\triangle ABC$ has base $AC=\sqrt6$ and height $\sqrt3-\frac{\sqrt6}{2},$ so its area is \[\frac12\cdot\sqrt6\cdot\left(\sqrt3-\frac{\sqrt6}{2}\right)=\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.\]
~MRENTHUSIASM