# 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 do I factorize/break it in terms of known expressions? Moreover, is there any general algorithm for breaking down these expressions into known lower-level powers since those formulae are pretty hard to remember and also lengthy to derive by trial and error?

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.

• Thats pretty neat! Apr 22, 2021 at 9:24

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:

$$a_3 = \alpha^3 + \beta^3 + \gamma^3 = a_1 - a_0 = -3$$ $$a_4 = \alpha^4 + \beta^4 + \gamma^4 = a_2 - a_1 = 2$$ $$\ldots$$