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 $ABC$?

Enter your answer in the form $\frac{a \sqrt{b} - c}{d},$ and simplified as usual.
We can translate the solutions, to obtain the equation $z^8 = 81 = 3^4.$  Thus, the solutions are of the form
\[z = \sqrt{3} \operatorname{cis} \frac{2 \pi k}{8},\]where $0 \le k \le 7.$  The solutions are equally spaced on the circle with radius $\sqrt{3},$ forming an octagon.

[asy]
unitsize(1 cm);

int i;

draw(Circle((0,0),sqrt(3)));
draw((-2,0)--(2,0));
draw((0,-2)--(0,2));

for (i = 0; i <= 7; ++i) {
  dot(sqrt(3)*dir(45*i));
  draw(sqrt(3)*dir(45*i)--sqrt(3)*dir(45*(i + 1)));
}

label("$\sqrt{3}$", (sqrt(3)/2,0), S);
[/asy]

We obtain the triangle with minimal area when the vertices are as close as possible to each other, so we take consecutive vertices of the octagon.   Thus, we can take $\left( \frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2} \right),$ $(\sqrt{3},0),$ and $\left( \frac{\sqrt{6}}{2}, -\frac{\sqrt{6}}{2} \right).$

[asy]
unitsize(1 cm);

int i;
pair A, B, C;

A = (sqrt(6)/2,sqrt(6)/2);
B = (sqrt(3),0);
C = (sqrt(6)/2,-sqrt(6)/2);

fill(A--B--C--cycle,gray(0.7));
draw(Circle((0,0),sqrt(3)));
draw((-2,0)--(2,0));
draw((0,-2)--(0,2));
draw(A--C);

for (i = 0; i <= 7; ++i) {
  dot(sqrt(3)*dir(45*i));
  draw(sqrt(3)*dir(45*i)--sqrt(3)*dir(45*(i + 1)));
}

label("$(\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2})$", A, A);
label("$(\sqrt{3},0)$", B, NE);
label("$(\frac{\sqrt{6}}{2}, -\frac{\sqrt{6}}{2})$", C, C);
[/asy]

The triangle has base $\sqrt{6}$ and height $\sqrt{3} - \frac{\sqrt{6}}{2},$ so its area is
\[\frac{1}{2} \cdot \sqrt{6} \cdot \left( \sqrt{3} - \frac{\sqrt{6}}{2} \right) = \boxed{\frac{3 \sqrt{2} - 3}{2}}.\]
We can translate the solutions, to obtain the equation $z^8 = 81 = 3^4.$  Thus, the solutions are of the form
\[z = \sqrt{3} \operatorname{cis} \frac{2 \pi k}{8},\]where $0 \le k \le 7.$  The solutions are equally spaced on the circle with radius $\sqrt{3},$ forming an octagon.

[asy]
unitsize(1 cm);

int i;

draw(Circle((0,0),sqrt(3)));
draw((-2,0)--(2,0));
draw((0,-2)--(0,2));

for (i = 0; i <= 7; ++i) {
  dot(sqrt(3)*dir(45*i));
  draw(sqrt(3)*dir(45*i)--sqrt(3)*dir(45*(i + 1)));
}

label("$\sqrt{3}$", (sqrt(3)/2,0), S);
[/asy]

We obtain the triangle with minimal area when the vertices are as close as possible to each other, so we take consecutive vertices of the octagon.   Thus, we can take $\left( \frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2} \right),$ $(\sqrt{3},0),$ and $\left( \frac{\sqrt{6}}{2}, -\frac{\sqrt{6}}{2} \right).$

[asy]
unitsize(1 cm);

int i;
pair A, B, C;

A = (sqrt(6)/2,sqrt(6)/2);
B = (sqrt(3),0);
C = (sqrt(6)/2,-sqrt(6)/2);

fill(A--B--C--cycle,gray(0.7));
draw(Circle((0,0),sqrt(3)));
draw((-2,0)--(2,0));
draw((0,-2)--(0,2));
draw(A--C);

for (i = 0; i <= 7; ++i) {
  dot(sqrt(3)*dir(45*i));
  draw(sqrt(3)*dir(45*i)--sqrt(3)*dir(45*(i + 1)));
}

label("$(\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2})$", A, A);
label("$(\sqrt{3},0)$", B, NE);
label("$(\frac{\sqrt{6}}{2}, -\frac{\sqrt{6}}{2})$", C, C);
[/asy]

The triangle has base $\sqrt{6}$ and height $\sqrt{3} - \frac{\sqrt{6}}{2},$ so its area is
\[\frac{1}{2} \cdot \sqrt{6} \cdot \left( \sqrt{3} - \frac{\sqrt{6}}{2} \right) = \boxed{\frac{3 \sqrt{2} - 3}{2}}.\]
The final answer is $$\frac{3 \sqrt{2} - 3}{2}$$. I hope it is correct.