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: 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
$$x^4+y^4+z^4=a_{4}=a_{3}+\dfrac{1}{2}a_{2}+\dfrac{1}{6}a_{1}=\dfrac{25}{6}$$
A: $x^4+y^4+z^4=\left(x^2+y^2+z^2\right)^2-2\left(x^2y^2+y^2z^2+z^2x^2\right)\tag{1}$
$\left(x^2y^2+y^2z^2+z^2x^2\right)  =(xy+yz+zx)^2-2xyz(x+y+z)\tag{2}$
Now note that 
$$\color{red}{\boxed{(xy+yz+zx)=\dfrac{1}{2}\left((x+y+z)^2-\left(x^2+y^2+z^2\right)\right)}}$$
from which you get the value of $xy+yz+zx$. Also, note that,
$$\color{blue}{\boxed{x^3+y^3+z^3-3xyz=(x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)}}$$
from which you get the value of $xyz$. 
Can you take it from here?
A: 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 $t^3-(1)t^2+bt-c=0$
$\implies t^4=t^3-bt^2+ct$
$\implies\sum x^4=\sum x^3-b\sum x^2+c\sum x$
A: 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
$$
  p_4=e_1p_3-e_2p_2+e_3p_4-4e_4=1\times3+\frac12\times2+\frac16\times1-0=\frac{25}6.
$$
A: 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.
