<s>[INST] 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? 
 [/INST] 
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.
</s>