<s>[INST] Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$

The polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$.
I am stuck here.The answer for $a+b+c+d=10$
 [/INST] You got stuck at a very odd point. You basically solved the hard part of the problem. So compute the are of the triangle, subtract it from the area of the octagon, and express the result in the desired form. 
Area of triangle: $\frac{1}{2}\sqrt{2}(1-\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}-1}{2}$. 
So the area of the polygon is $2\sqrt{2}- \frac{\sqrt{2}-1}{2}= \frac{3\sqrt{2}+1}{2}$. 
The result is not unique though. Given any number $t$, $a=3t, b=2, c=t, d=2t$ is a solution (and there are many others). So $a+b+c+d= 6t+2$ can be any number. 
Some condition must be missing, check the problem again.
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