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 $\alpha^6+\beta^6$ ?
 A: 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 $u_6$. 
A: Hint: Here $\alpha^2=2\alpha -4$ and same for $\beta$, as they are the roots of the equation $x^2-2x+4=0$.
Now calculate the value of ${\alpha}^6$ and ${\beta}^6$ in terms of $\alpha^2 and \beta^2$. Then substitute their value again.
A: Use this one:
$$[(a+b)^2-2ab)][((a+b)^2-2ab)^2-3a^2b^2]$$
A: Well, in general:
$$\text{a}\cdot x^2+\text{b}\cdot x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag1$$
So, for your example we can set:


*

*$$\alpha=x_+=\frac{-\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag2$$

*$$\beta=x_-=\frac{-\text{b}-\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag3$$


Now, for your problem:
$$\alpha^6+\beta^6=\left(\frac{-\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\right)^6+\left(\frac{-\text{b}-\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\right)^6=$$
$$\frac{\left(-\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}\right)^6+\left(\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}\right)^6}{64\cdot\text{a}^6}\tag4$$
