<s>[INST] Polynomial system If there are 3 numbers $x,y,z$ satisfying
$f=x+y+z=3$ ,  $g=x^2+y^2+z^2=5$ , $h=x^3+y^3+z^3=7$ then prove that they also satisfy 
$x^4+y^4+z^4=9$ but not $x^5+y^5+z^5=11$
I dont know how to tackle this to be honest, i have started trying to write $x^4+y^4+z^4-9$ as $r_1\times f + r_2\times g+ r_3\times h$ where $r_1,r_2,r_3$ in $C[x,y,z]$ but then i realised that i cant possibly do that by hand.This was an exercise on my first computational algebra course so we havent really learnt anything much yet. I would like a hint if possible.
 [/INST] $$2(xy+yz+zx) = (x+y+z)^2 - (x^2+y^2+z^2) = 4$$
$$-3xyz = (x+y+z)^3 - (x^3+y^3+z^3) - 3(xy+yz+zx)(x+y+z) = 27 - 25 = 2$$
Thus $x,y,z$ are the roots of 
$$t^3 -3t^2 + 2t + \frac{2}{3} = 0$$ 
Now
\begin{align*}
t^4 = 3t^3 -2t^2 -\frac{2}{3}t
\end{align*}
Putting $t=x,y,z$ in the above and adding, we get
$$x^4 + y^4 + z^4 = 3 \sum x^3 - 2\sum x^2 - \frac{2}{3} \sum x = 21 - 10 -2 = 9$$
Also,
$$t^5 = 3t^4 - 2t^3 - \frac{2}{3} t^2$$ and hence
$$\sum x^5 = 3\sum x^4 - 2 \sum x^3 - \frac{2}{3} \sum x^2 = 27 - 14 - 10/3 = 29/3$$
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