<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] Hint:
let $a_{n}=x^n+y^n+z^n$
then we have
$$a_{n+2}=(x+y+z)a_{n+1}-(xy+yz+xz)a_{n}+xyza_{n-1}$$
it is easy to find
$$xy+yz+xz=\dfrac{1}{2}[(x+y+z)^2-(x^2+y^2+z^2)]=-\dfrac{1}{2}$$
$$x^3+y^3+z^3-3xyz=(x+y+z)^3-3(xy+yz+xz)(x+y+z)$$
then we have
$$xyz=\dfrac{1}{6}$$
so
$$a_{n+2}=a_{n+1}+\dfrac{1}{2}a_{n}+\dfrac{1}{6}a_{n-1}$$
so
$$x^4+y^4+z^4=a_{4}=a_{3}+\dfrac{1}{2}a_{2}+\dfrac{1}{6}a_{1}=\dfrac{25}{6}$$
</s>