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 $x^4+y^4+z^4=?$

A key to this kind of question is a formed by Newton's identities, which relate the power sums $p_k=\sum_{i=1}^nx^i$ for different values of $k$ to the elementary symmetric polynomials $e_k$; one can define $e_k$ as the coefficient of $X^k$ in the expansion of $\prod_{i=1}^n(1+x_iX)$. The relations are most easily stated recursively as
$$
  ke_k=\sum_{i=1}^k(-1)^{i-1}p_ie_{k-i}
$$
(note that the final term involves $e_0=1$). This makes converting between values $p_1,p_2,\ldots,p_k$ and values $e_1,e_2,\ldots,e_k$ easy.
The problem actually gives $p_k=k$ for $k=1,2,3$, and asks for $p_4$, which normally Newton's identities cannot do. However the additional fact that makes this doable is that one has $n=3$ here: with only three values $x_i$, we know that $e_k=0$ for all $k>3$, in particular for $k=4$.
So first solve $e_1,e_2,e_3$ from
$$ \begin{align} e_1&=p_1e_0=1 \\ 2e_2&=p_1e_1-p_2e_0=e_1-2 \\
 3e_3&=p_1e_2-p_2e_1+p_3e_0=e_2+2e_1+3
  \end{align}
$$
giving $(e_1,e_2,e_3)=(1,-\frac12,\frac16)$, and then use $e_4=0$ in the reversed identity
$$
  p_4=e_1p_3-e_2p_2+e_3p_4-4e_4=1\times3+\frac12\times2+\frac16\times1-0=\frac{25}6.
$$
