1,2,3,will this continue?

Relevant wiki: Newton's Identities

Let $P_n = a^n+b^n+c^n$ , where $n$ is a positive integer, and $S_1 = a+b+c=1$ , $S_2 = ab+bc+ca$ , and $S_3 = abc$ . By Newton's sums or Newton's identities:

$\begin{aligned} P_1 & = S_1 = 1 \\ P_2 & = S_1P_1 - 2S_2 = 1(1) - 2S_2 = 2 & \small \color{#3D99F6} \implies S_2 = - \frac 12 \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = 1(2) + \frac 12 (1) + 3S_3 = 3 & \small \color{#3D99F6} \implies S_3 = \frac 16 \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 = 1(3) + \frac 12 (2) +\frac 16(1) = 4\frac 16 \end{aligned}$

Therefore, $P_4 - 4 = \boxed{\dfrac 16}$ .

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 $P_4=\frac{4*3*1}{3}-2*{1}^2+\frac{{1}^4}{6}+\frac{{2}^2}{2} = 4 + \frac{1}{6}.$