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

$x^4+y^4+z^4=\left(x^2+y^2+z^2\right)^2-2\left(x^2y^2+y^2z^2+z^2x^2\right)\tag{1}$
$\left(x^2y^2+y^2z^2+z^2x^2\right)  =(xy+yz+zx)^2-2xyz(x+y+z)\tag{2}$
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?
