12,12,12,will this continue?

We can find the solution of x^4 + y^4 + z^4 by the following steps: (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+xz+yz), so that (xy+xz+yz) = 66, (x+y+z)^3 = x^3 + y^3 + z^3 + 3(xy+xz+yz) (x+y+z) – 3xyx , then we find :xyz = 220, (xy+xz+yz)^2 = (xy)^2 + (xz)^2 + (yz)^2 + 2xyz(x+y+z), then we can find (xy)^2 + (xz)^2 + (yz)^2 = - 924, (x^2 + y^2 + z^2)^2 = x^4 + y^4 + z^4 + 2((xy)^2 + (xz)^2 + (yz)^2), finally we can find x^4 + y^4 + z^4 = 1992

Let $P_n = x^n + y^n+z^n$ , where $n$ is a positive integer, $S_1=x+y+z=12$ , $S_2 = xy + yz+zx$ , and $S_3 = xyz$ . Given $P_1=P_2=P_3=12$ , we need to find $P_4 = x^4+y^4+z^4$ . By Newton's sums or Newton's identities, we have:

$\begin{aligned} P_1 & = S_1 = 12 \\ P_2 & = S_1P_1 - 2S_2 = 12(12) -2S_2 = 12 & \small \color{#3D99F6} \implies S_2 = 66 \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = 12(12) -66(12) + 3S_3 = 12 & \small \color{#3D99F6} \implies S_3 = 220 \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 = 12(12) -66(12) + 220(12) = \boxed{1992} \end{aligned}$

$\left. \begin{aligned} 144&=(x+y+z)^2\\ &=(x^2+y^2+z^2)+2(xy+xz+yz)\\ &=12+2(xy+xz+yz) \end{aligned} \right\} \Rightarrow xy+xz+yz=66\\$

$\left. \begin{aligned} 792&=(xy+xz+yz)(x+y+z)\\ &=3xyz+[x^2(y+z)+y^2(x+z)+z^2(x+y)]\\ &=3xyz+[x^2(12-x)+y^2(12-y)+z^2(12-z)]\\ &=3xyz+132 \end{aligned} \right\} \Rightarrow xyz=220\\$

$\left. \begin{aligned} 144&=(x^3+y^3+z^3)(x+y+z)\\ &=(x^4+y^4+z^4)+[xy(x^2+y^2)+xz(x^2+z^2)+yz(y^2+z^2)]\\ &=(x^4+y^4+z^4)+[xy(12-z^2)+xz(12-y^2)+yz(12-x^2)]\\ &=(x^4+y^4+z^4)+12(xy+xz+yz)-xyz(x+y+z)\\ &=(x^4+y^4+z^4)+792-2640 \end{aligned} \right\} \Rightarrow x^4+y^4+z^4=1992\\$

$x^4+y^4+z^4=-\frac{2}{3} \left(x^3+y^3+z^3\right) (x+y+z)+2 \left(x^3+y^3+z^3\right) (x+y+z)-\left(x^2+y^2+z^2\right) (x+y+z)^2-\frac{1}{2} \left(x^2+y^2+z^2\right)^2+\left(x^2+y^2+z^2\right)^2+\frac {1}{6} (x+y+z)^4=1992$

My workings isn't as short or as elegant as those that had already posted, but this was how I worked it out.

Using Vieta's formulas, where $x, y, z$ are the roots of a cubic polynomial:

$\sum x = 12$

$(x + y +z)^2 = x^2 + y^2 + z^2 + 2\sum xy \implies \sum xy = 66$

Let $S_n = x^n + y^n + z^n$ :

Then $S_3 - 12S_2+66S_1 - 3c = 0 \implies c = 220$

So the equation is $w^3 - 12w^2 + 66w - 220 = 0$

Now we can solve for $S_4 = x^4 + y^4 + z^4$ , by substituting the given equations in the question.

$S_4 - 12S_3 + 66S_2 - 220S_1 = 0$

$S_4 = 12(12) - 66(12) + 220(12) = 1992$

$144 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) = 12 + 2(xy + xz + yz) \implies \boxed{xy + xz + yz = 66}$ .

$1728 = (x + y + z)^3 = x^3 + y^3 + z^3 + 3x^2y + 3xy^2 + 3x^2z + 6xyz + 3y^2z + 3xz^2 + 3yz^2 =$ $x^3 + y^3 + z^3 + 3(x + y + z)(xy + xz + yz) - 3xyz = 12 + 2376 - 3xyz = 2388 - 3xyz \implies \boxed{xyz = 220}$ .

$(20736 = (x + y + z)^4 = x^4 + y^4 + z^4 + (4x^3y + 4xy^3 + 4x^3z + 4y^3z + 4xz^3 + 4yz^3) +$

$(6x^2y^2 + 12x^2yz + 12xy^2z + 6x^2z^2 + 12xyz^2 + 6y^2z^2)$ .

$6(xy + xz + yz)^2 =6x^2y^2 + 12x^2yz + 12xy^2z + 6x^2z^2 + 12xyz^2 + 6y^2z^2$ .

and,

$4(x^2 + y^2 + z^2)(xy + xz + yz) = 4x^3y + 4xy^3 + 4x^3z + 4y^3z + 4xz^3 + 4yz^3 + 4(xyz)(x + y + z) \implies$

$4x^3y + 4xy^3 + 4x^3z + 4y^3z + 4xz^3 + 4yz^3 = 4(x^2 + y^2 + z^2)(xy + xz + yz) - 4(xyz)(x + y + z)$

$\implies (x + y + z)^4 = 20736 = x^4 + y^4 + z^4 + 6(66^2) + 4(12)(66) - 4(220)(12) = x^4 + y^4 + z^4 + 18744 \implies$

$\boxed{x^4 + y^4 + z^4 = 1992}$ .

If $(P_i=x^i+y^i+z^i)$

According to Newton's identities ,

$P_4=\frac{4P_3P_1}{3}-P_2{P_1}^2+\frac{{P_1}^4}{6}+\frac{{P_2}^2}{2}$

so we get $P_4=1992.$