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? 
 A: Let
$$p_i=x^i+y^i+z^i$$
and let
$$e_1=x+y+z$$
$$e_2=xy+xz+yz$$
and
$$e_3=xyz$$
By Newton's identities,
$$e_1p_3-e_2p_2+e_3p_1=p_4$$
We also have
$$e_1=p_1$$
$$2e_2=e_1p_1-p_2=p_1^2-p_2=132$$
so that
$$e_2=66$$
Also
$$3e_3=e_2p_1-e_1p_2+p_3=66\cdot 12-12\cdot 12+12=660$$
so that
$$e_3=220$$
Hence
$$p_4=12\cdot 12-12\cdot 66+12\cdot 220= 1992$$
A: Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, $$u=4,$$
$$3v^2=\frac{12^2-12}{2}=66,$$ which gives $$v^2=22.$$
Also, $$12=x^3+y^3+z^3=27u^3-27uv^2+3w^3,$$ which gives $$w^3=220$$ and use
$$x^4+y^4+z^4=81u^4-108u^2v^2+18v^4+12uw^3.$$
I got $1992.$
I used the known $uvw$'s substitutions: https://artofproblemsolving.com/community/c6h278791
A: 
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.
A: If you call $S_i=\sum x^i+y^i+z^i$ then we have this relation:
$$S_4=\frac 16\left({S_1}^4+8\,S_3\,S_1-6\,S_2\,{S_1}^2+3\,{S_2}^2\right)$$
You can have a look at some proofs in this topic: Find the value of $a^4+b^4+c^4$
A: The following method is due to Euler, also explained here. Let's write:
$$S_r = x_1^r + x_2^r +x_3^r$$
for $r\in \mathbb{N}$ 
Consider the function:
$$f(u) = -\sum_{j=1}^3\log\left(1-\frac{x_j}{u}\right)$$
The series expansion around infinity of this function is given by:
$$f(u) = \sum_{k=1}^{\infty}\frac{S_k}{k u^k}\tag{1}$$
We also have that $\exp\left[-f(u)\right]$ is a third degree polynomial in $u^{-1}$. This means that we can compute $S_4$  by exponentiating the series (1), computing the fourth order term in $u^{-1}$ and setting the result equal to zero.
In this particular case we can simplify this, by writing:
$$f(u) = 12 \sum_{k=1}^{\infty}\frac{1}{k u^k} + \sum_{k=4}^{\infty}\frac{S_k'}{k u^k} = -12 \log\left(1-\frac{1}{u}\right) +  \sum_{k=4}^{\infty}\frac{S_k'}{k u^k}$$
where $S'_k = S_k - 12$.
So, we have:
$$\exp\left[-f(u)\right] = \left(1-u^{-1}\right)^{12}\left[1- \frac{S_4 - 12}{4 u^4}\right] +\mathcal{O}(u^{-5})$$
This being a polynomial of 3rd degree in $u^{-1}$, means that the coefficient of $u^{-4}$ equals zero. We thus have:
$$S_4 = 4 \left[\binom{12}{4} + 3\right] = 1992$$
A: The following method, using Cayley–Hamilton theorem, can be seen as using the Newton identities secretly, but is more transparent in my opinion:
Let $$ X =     \left(\begin{matrix}
    x & 0 & 0 \\
    0 & y & 0 \\
    0 & 0 & z \\
    \end{matrix} \right) \,.$$
We are interested in traces of $X^n$. The characteristic polynomial of $X$ is
$$p(\lambda) = \det(\lambda - X) = \lambda^3 - (x+y+z) \lambda^2 + (xy + yz + zx)\lambda - xyz\,.$$
The second coefficient is $-12$, the third coefficient can be found by calculating $(x+y+z)^2 - (x^2 + y^2 + z^2)$, giving $66$
To find $xyz$, one can either do algebra as in other answers, or use the Cayley-Hamilton theorem, which says that $p(X) = 0$. Taking trace, one gets $3xyz = 12 - 12\cdot 12+12\cdot 66$, or $xyz = 220$. 
Thus (from Cayley-Hamilton)
$$ p(X) = X^3 - 12X^2 + 66X - 220 = 0\,.$$
From this identity, it's easy to obtain the recursion relation
$$S_n =  x^n + y^n + z^n = \mathrm{tr} [X^n] = \mathrm{tr}[X^{n-3}  (12 X^2 - 66X + 220) ]  = 12 S_{n-1} - 66 S_{n-2} + 220 S_{n-3}$$
giving 
$$S_4 = (12-66+220)\cdot 12 = 1992$$
