Linear system of equations that arises from recurrence relation. So when solving a third order homogenous recurrence relation I end upp with the task of determening the constants $a,b$ and $c$. The initial conditions give rise to the following system of equations:
$$\left\{
  \begin{array}{rcr}
    2a+(1+\sqrt{6})b+(1-\sqrt{6})c & = & 0\\
    4a+(1+\sqrt{6})^2b+(1-\sqrt{6})^2c & = & 0\\
    8a+(1+\sqrt{6})^3b+(1-\sqrt{6})^3c & = & 1
  \end{array}
\right.$$
I've verified that this system is correct but now I need to solve it. I find brute force Gaussian elimination quite tedious to do (or maybe there is some trick to make it simple?). Is there any other easier method to solve this without having to spend 3-4 pages of hand written scribble?
The answer is
$$a=-\frac{1}{10}, \ b=\frac{1}{10\sqrt{6}}, \ c=-\frac{1}{10\sqrt{6}}.$$

Very basic (done by hand in less than a page with pen and paper).
Consider 
   $$2 a+A b+B c  = 0\tag 1$$
$$ 4 a+A^2 b+B^2 c =0\tag 2$$
$$ 8 a+A^3 b+B^3 c =1\tag 3$$ 
Use $(1)$ and $(2)$ and solve for $(a,b)$ to get 
$$a=\frac{B  (B-A)}{2 (A-2)} c\qquad \text{and} \qquad b=\frac{(2-B) B }{(A-2) A}c$$ Plug in $(3)$ and solve for $c$ to get
$$c=\frac{1}{(2-B) B (A-B)}\implies a=\frac{1}{2 (A-2) (B-2)}\qquad \text{and} \qquad b=\frac{1}{(A-2) A (A-B)}$$
Now, make $A=1+\sqrt 6$ and $B=1-\sqrt 6$.
