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.

For notational ease, let's define a few variables:
\begin{align*}
j&=x^3y+x^3z+xy^3+xz^3+y^3z+yz^3\\
k&=x^2y^2+x^2z^2+y^2z^2\\
l&=x^2yz+xy^2z+xyz^2.
\end{align*}
Now, let's start multiplying things out:
\begin{align*}
fh&=(x^4+y^4+z^4)+(xy^3+xz^3+yx^3+yz^3+zx^3+zy^3)\\
&=(x^4+y^4+z^4)+j\\
g^2&=(x^4+y^4+z^4)+2(x^2y^2+x^2z^2+y^2z^2)\\
&=(x^4+y^4+z^4)+2k\\
f^2g&=((x^2+y^2+z^2)+2(xy+xz+yz))(x^2+y^2+z^2)\\
&=(x^4+y^4+z^4)+2(x^2y^2+x^2z^2+y^2z^2)\\
&\qquad\qquad+2(x^3y+x^3z+x^2yz+xy^3+xy^2z+y^3z+xyz^2+xz^3+yz^3)\\
&=(x^4+y^4+z^4)+2j+2k+2l\\
f^4&=(x^4+y^4+z^4)+4(xy^3+xz^3+yx^3+yz^3+zx^3+zy^3)+6(x^2y^2+x^2z^2+y^2z^2)\\
&\qquad\qquad+12(x^3yz+xy^3z+xyz^3)\\
&=(x^4+y^4+z^4)+4j+6k+12l
\end{align*}
Combining, we have
\begin{align*}
21&=fh=(x^4+y^4+z^4)+j\\
25&=g^2=(x^4+y^4+z^4)+2k\\
45&=f^2g=(x^4+y^4+z^4)+2j+2k+2l\\
81&=f^4=(x^4+y^4+z^4)+4j+6k+12l
\end{align*}
This is now a system of equations where we can use linear algebra to solve.  For example, subtracting $6$ times the third from the fourth gives
$$
-189=-5(x^4+y^4+z^4)-8j-6k
$$
Now, adding $3$ times the second to this gives
$$
-114=-2(x^4+y^4+z^4)-8j
$$
Finally, adding $8$ times the first to this gives
$$
54=6(x^4+y^4+z^4)
$$
In other words, $x^4+y^4+z^4=9$.
