Can I solve this other than using Newton's sums or Newton's identities? 
$\begin{cases}
 &x + y + z = 12\\& x^2 + y^2 + z^2 =  12
\\ & x^3 +y^3 + z^3 = 12
\end{cases}$
If $x,$ $y,$ and $z$ satisfy the system of equations above, what is the value of $x^4+y^4+z^4?$ 

People told me that this can be solved by using Newton's sums or Newton's identities which I don't know how. Does anybody know how to do that or any other method to solve this? 


People told me that this can be solved by using Newton's sums or Newton's identities which I don't know how. Does anybody know how to do that or any other method to solve this?

We can solve this by doing this method:
$\begin{align} (x+y+z)^2 &\rightarrow x^2 + y^2 + z^2 + 2(xy+xz+yz) = 12^2 \\
&\rightarrow 12 + 2(xy+xz+yz) = 144 \\
& \rightarrow 2(xy+xz+yz) = 132 \\
& \rightarrow (xy+xz+yz) = 66\end{align}$
$\begin{align} (x+y+z)^3 &\rightarrow x^3 + y^3 + z^3 + 3(xy+xz+yz)(x+y+z) – 3xyz = 12^3 \\
&\rightarrow 12 + 3(66)(12) – 3xyz = 1728 \\
&\rightarrow 2,388 – 3xyz = 1728 \\
&\rightarrow – 3xyz = -660 \\
& \rightarrow xyz = 220 \end{align}$
$\begin{align} (xy+xz+yz)^2 & \rightarrow (xy)^2 + (xz)^2 + (yz)^2 + 2xyz(x+y+z) = 66^2 \\
& \rightarrow (xy)^2 + (xz)^2 + (yz)^2 + 2xyz(x+y+z) = 66^2 \\
& \rightarrow (xy)^2 + (xz)^2 + (yz)^2 + 2(220)(12) = 4356 \\
& \rightarrow (xy)^2 + (xz)^2 + (yz)^2 + 5280 = 4356 \\
& \rightarrow (xy)^2 + (xz)^2 + (yz)^2  = -924 \end{align}$
$\begin{align} (x^2 + y^2 + z^2)^2 & \rightarrow x^4 + y^4 + z^4 + 2\left( (xy)^2 + (xz)^2 + (yz)^2 \right) = 12^2 \\
&\rightarrow x^4 + y^4 + z^4 + 2\left( -924 \right) = 144 \\
&\rightarrow x^4 + y^4 + z^4 + (-1848) = 144 \\
&\rightarrow x^4 + y^4 + z^4 + (-1848) = 144 \\
& \rightarrow x^4 + y^4 + z^4 = \boxed{1992} \end{align}$
Obviously we are deaing with complex numbers here, $x = 2.467+5.005i,$ $y = 7.066,$ $z = 2.467-5.005i$ is one solution.
