How does recurrence relation works with system of equations? There are      $$   a+b+c+d = 2\\2a+2^2b+2^3c+2^4d = 5\\ 3a+3^2b+3^3c+3^4d = 6\\4a+4^2b+4^3c+4^4d = 1$$
then I'm given 
$$C_{n}= a+bn+cn^2+dn^3$$
from linear recurrence relation with repeated roots said that

$$(x-1)^4 = \sum_{k=0}^{4}\binom{4}{k}(-1)^{4-k}x^{4-k}$$

and then   
$$C_{n}=\sum_{k=1}^{4}\binom{4}{k}(-1)^{5-k}C_{n-k} = 4C_{n-1}-6C_{n-2}+4C_{n-3}-C_{n-4}$$ $\forall n \geq 5$
I want to find $C_{6}$. Then i've plugged $C_{5}=-3$ into that relation which finally gave $C_{6}=-8$ but wolfram alpha told me that $C_{6} =-48 
Question: How do I use recurrence relations to find out the correct $C_6$

I believe you are over-complicating it. The given system tells you that the quartic polynomial $p(n)=an+bn^2+cn^3+dn^4$ attains the values $2,5,6,1$ at $n=1,2,3,4$. By Lagrange interpolation such polynomial is 
$$ p(n) = \frac{1}{24}\left(18n+37n^2-6n^3-n^4\right) $$
hence
$$ C_6 = \frac{p(6)}{6} = \color{red}{-8}.$$
