Given $n$ equations in cyclic format how to solve the system? If $x_1+x_2+x_3 = c_1$
$x_2+x_3+x_4 = c_2$
$x_3+x_4+x_5 = c_3$
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$x_{99}+x_{100}+x_1 = c_{99}$
$x_{100}+x_1+x_2 = c_{100}$
$x_i$ denote variables and $c_i$ denote constants.
How to calculate $x_1,x_2,\dots,x_{100}?$
 A: Edit: this answer was largerly edited, but keeping the same methodology, dealing with this problem by pure polynomial manipulation
Let us call $D$ the cyclic delay operator, the matrix is equal to $A=I+D+D^2$ with $D^{100}=I$
All circulant matrices can be represented in this way. A key point here is that the subset (ring) of circulant matrices is commutative, which allows classical algebric manipulations, here formal polynomial manipulation.
We first note that 
$$\sum_{i=0}^{99} D^i = (I+D+D^2) \sum_{i=0}^{32}D^{3i} + D^{99}$$
$$ D^{99} = D^{-1}$$
$$(I+D+D^2) \sum_{i=0}^{99} D^i = 3\, \sum_{i=0}^{99} D^i $$
It follows:
$$(I+D+D^2)^{-1} = \frac{1}{3} \sum_{i=0}^{99} D^i - \sum_{i=0}^{32} D^{3i+1}$$
And finally:
$$x_j = \frac{1}{3} \sum_{i=1}^{100} c_i - \sum_{i=0}^{32} c_{j+3i+1}   $$
When the indices $j+3i+1$ must be understood in a 'cyclic' way.
Note 1: the matrix $A$ corresponds to a cyclic filtering operation. The inversion can be interpreted as equalization
Note 2: all circulant matrices have same eigenvectors, corresponding to the DFT matrix. This can be used for example to show that the corresponding ring is commutative. It could be used also to perform the matrix inversion ("equalization in the frequency domain"). This proves that when an inverse of a circulant matrix exists, this matrix is circulant
A: I inverted the matrix of coefficients for the $10 \times 10$ case in a spreadsheet.  If we call the entries of the matrix $a_{ij}$ we get $$a_{ij}=\begin {cases} -\frac 23&j \gt i, j-i \equiv 1 \pmod 3 \\ -\frac 23 & i \gt j,i=j \equiv 0 \pmod 3 \\\frac 13 & \text {otherwise} \end {cases}$$
You can multiply the column matrix of the $c_i$ by this to get the $x_i$.  I am sure the pattern will continue for all matrices of size $1 \bmod 3$.  You can verify the inverse by multiplying it by your coefficient matrix.  It will take some keeping track of indices, but it should work.  An image of three times the inverse in the $10 \times 10$ case is below.
 
This gives $$x_1=\frac 13 \sum_{i=1}^{100} c_i-\sum_{j=1}^{33}c_{3j-1}$$ and all the others are cyclic shifts of this.
