A Curious Problem

We can solve this type of problem with Newton's sums (identities) . First let $P_n = x^n+y^n+z^n$ , where $n$ is a positive integer. And $S_1 = x+y+z$ , $S_2 = xy+yz +zx$ , and $S_3 = xyz$ . Then we have:

$\begin{aligned} P_1 & = S_1 = 1 & \small \blue{\implies S_1 = 1} \\ P_2 & = S_1P_1 - 2S_2 = 1 -2S_2 = 2 & \small \blue{\implies S_2 = - \frac 12} \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = 2 + \frac 12 + 3S_3 = 3 & \small \blue{\implies S_3 = - \frac 16} \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 = 3 + 1 + \frac 16 = \frac {25}6 \approx \boxed{4.167} \end{aligned}$

$2(xy+yz+zx)=(x+y+z)^{2}-(x^{2}+y^{2}+z^{2})=-1$

$xy+yz+zx=1/2$

$6xyz=(x+y+z)^{3}-3(x+y+z)(x^{2}+y^{2}+z^{2})+2(x^{3}+y^{3}+z^{3})=1$

$xyz=1/6$

$x^{4}+y^{4}+z^{4}=(x^{2}+y^{2}+z^{2})^{2}-2(x^{2}y^{2}+y^{2}x^{2}+z^{2}x^{2})$

$=(x^{2}+y^{2}+z^{2})^{2}-2((xy+yz+zx)^{2}-2xyz(x+y+z))$

$=2^{2}-2((-1/2)^{2}-2(1/6)(1))$

$=4-2(1/4-1/3)$

$=4+2/12$

$=4+1/6$

$=25/6$

$=4.1666666666666...7$