General deduction of expression $x^n + y^n + z^n $ provided some of its previous powers are known. 
If $x^3 -x +1 = 0$ and $\alpha$ , $\beta$ , $\gamma$ are its roots ,find $\alpha^2+\beta^2+\gamma^2$  ; $\alpha^3+\beta^3+\gamma^3$ ; $\alpha^4+\beta^4+\gamma^4$ and $\alpha^5+\beta^5+\gamma^5$

My approach:
1. Find $\alpha^2+\beta^2+\gamma^2$
$$\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma + \gamma\alpha)$$
Putting the values we get $$\alpha^2+\beta^2+\gamma^2 = 2$$
2. Find $\alpha^3+\beta^3+\gamma^3$
$$\alpha^3+\beta^3+\gamma^3 = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma - \gamma\alpha) + 3\alpha\beta\gamma$$
Here we substitute the value of $\alpha^2+\beta^2+\gamma^2$ and get $\alpha^3+\beta^3+\gamma^3=3$
3. Find $\alpha^4+\beta^4+\gamma^4$
$$\alpha^4+\beta^4+\gamma^4 = (\alpha^2+\beta^2+\gamma^2)^2 -2[(\alpha\beta+\beta\gamma + \gamma\alpha)^2 -2\alpha\beta\gamma(\alpha+\beta+\gamma)]$$
Here we substitute the value of $\alpha^2+\beta^2+\gamma^2$ and get $\alpha^4+\beta^4+\gamma^4=2$
But I am stuck at $\alpha^5+\beta^5+\gamma^5$. How to factorize/break it in terms of known expressions?
And is there any general algorithm for breaking down these expressions into known lower level powers since those formualae are pretty hard to remember and also lengthy to  derive by trial and error ?
 A: Multiply the equation by $x^2$:
$$x^5-x^3+x^2=0$$
and plug $\alpha, \beta, \gamma$:
$$\begin{cases}\alpha^5-\alpha^3 +\alpha^2=0\\ \beta^5-\beta^3+\beta^2=0\\ \gamma^5-\gamma^3+\gamma^2=0\\ \end{cases}$$
and add them:
$$\alpha^5+\beta^5+\gamma^5=(\alpha^3+\beta^3+\gamma^3)-(\alpha^2+\beta^2+\gamma^2)=3-2=1.$$
Note: To find $x^3+y^3+z^3$, you can use the same trick.
A: There is a quite direct way using linear recurrencies.
$x^3-x+1$ is the characteristic polynomial of the linear recurrence
$$a_{n+3} - a_{n+1}+a_n=0$$
This has the general solution
$$a_n = A\alpha^n + B\beta^n + C\gamma^n \text{ for } n \geq 0$$
You are looking for the special solution where
$$A = B = C = 1$$
Now, you have
$$a_0 = A+B+C = 3$$
$$a_1 \stackrel{Vieta}{=} \alpha+\beta+\gamma = 0$$
$$a_2 = a_1^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \stackrel{Vieta}{=} 0-2\cdot(-1) = 2$$
Now, you can start the recursion:
$$a_3 = \alpha^3 + \beta^3 + \gamma^3 = a_1 - a_0 = -3$$
$$a_4 = \alpha^4 + \beta^4 + \gamma^4 = a_2 - a_1 = 2$$
$$ \ldots $$
