If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value of $\alpha^6+\beta^6$ is 
If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value
  of $\alpha^6+\beta^6$ is

I know that, here, $\alpha\beta=4$ and $\alpha + \beta = 2$ and use that result to find $\alpha^2 + \beta^2$ using the expansion of $(a+b)^2$
But how to find $\alpha^6+\beta^6$ ?

From $\alpha^2-2\alpha+4=0$ we get that for $n\geq 0$ we have $\alpha^{n+2}-2\alpha^{n+1}+4\alpha^n=0$, and we have the same relation for $\beta$. Putting $u_n=\alpha^n+\beta^n$ and adding, we get that $u_{n+2}-2u_{n+1}+4u_n=0$ for all $n$. We have $u_0=2$, $u_1=2$, and now it is easy to compute $u_6$. 
