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I had seen a fun problem that is exactly the problem in the question, except it was a specific case of this. Turns out, if $x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=3,$ then $x^5+y^5+z^5 = 6$. I put this equation into Wolfram Alpha for different values of $n$. For instance, $x^8+y^8+z^8=\frac{51}{72}$. $x^{11}+y^{11}+z^{11}=\frac{11117}{216}.$

Firstly, is there an explicit expression to evaluate $x^n+y^n+z^n$? And, if not, are we able to prove that this is always a rational number, at least? WolframAlpha is unable to calculate an explicit form for $x^n+y^n+z^n$.

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None of the calculations we need are mysterious, but one might need some experience dealing with symmetric functions and their relations to know what to look for.

We denote $x^n+y^n+z^n$ by $p_n$. You can verify the following inductive formula:

$p_n = (x+y+z)p_{n-1}-(xy+yz+zx)p_{n-2}+(xyz)p_{n-3}$

We can calculate the coefficients of the $p_i$ above as follows:

$x+y+z = 1$

$xy + yz + zx = \frac{(x+y+z)^2 -(x^2+y^2+z^2)}{2} = \frac{1^2-2}{2} = -\frac{1}{2}$

$xyz = \frac{(x^3+y^3+z^3)-(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}{3} = \frac{3-1 \cdot (2-(-\frac{1}{2}))}{3} = \frac{1}{6}$

Thus our inductive formula becomes

$p_n = p_{n-1}+\frac{1}{2}p_{n-2}+\frac{1}{6}p_{n-3}$

It is clear from this recursive equation that all of the $p_n$ will be rational numbers, since $p_1, p_2$ and $p_3$ are rational numbers. In fact, it even follows that the denominator of $p_n$ will be of the form $2^a3^b$ for some non-negative integers $a$ and $b$. For an explicit form we could look at roots of the characteristic/auxiliary polynomial of the corresponding recursive equation. In this case we get

$T^3-T^2-\frac{1}{2}T-\frac{1}{6}$

whose roots are not that nice. However, Wolfram Alpha gives an explicit formula. Then again, one could perhaps just solve the original system directly to find $x$, $y$ and $z$.

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Yes. In https://en.m.wikipedia.org/wiki/Symmetric_polynomial, look at Power-sum Symmetric Polynomials in the section Special kinds of symmetric polynomials

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