<s>[INST] If we know $x+y+z=1$, $x^2+y^2+z^2=2$, and $x^3+y^3+z^3=3$, how to find $x^4+y^4+z^4$? Let $x$, $y$, and $z$ be such that
$$\begin{align*}
x+y+z&=1\\
 x^2+y^2+z^2&=2 \\
 x^3+y^3+z^3&=3 
\end{align*}$$
Then $x^4+y^4+z^4=?$
 [/INST] An additional solution is using Groebner bases.  For this problem, consider the ring $R=\mathbb{Q}[x,y,z]$ and the ideal 
$$
I=\langle x+y+z-1,x^2+y^2+z^2-2,x^3+y^3+z^3-3\rangle.
$$
Using a computer algebra system we can find a Groebner basis (with the grRevLex order) as
$$
\{x+y+z-1,2y^2+2yz+2z^2-2y-2z-1,6z^3-6z^2-3z-1\}
$$
Then, computing the remainder of $x^4+y^4+z^4$ under division by this basis gives
$$
\frac{25}{6},
$$
as others have found through more elementary means.
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