Can I solve this other than using Newton's sums or Newton's identities?

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 $$p_4=12\cdot 12-12\cdot 66+12\cdot 220= 1992$$

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

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.