<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] We can use $$(x+y+z)^2=2\sum xy+(x^2+y^2+z^2)\ \ \ \ (1)$$
$$\sum x^3-3xyz=(x+y+z)(\sum x^2-\sum yz)$$
$$\iff\sum x^3-3xyz=(x+y+z)\{(x+y+z)^2-3\sum xy\}\ \ \ \ (2)$$
to find $xyz=c$(say)$,\sum xy=b$(say)
Then $x,y,z$ are the roots of $t^3-(1)t^2+bt-c=0$
$\implies t^4=t^3-bt^2+ct$
$\implies\sum x^4=\sum x^3-b\sum x^2+c\sum x$
</s>