An algebra problem by Nahom Assefa

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

$\begin{aligned} P_1 & = S_1 = 1 & \small \blue{\text{Given}} \\ P_2 & = S_1P_1 - 2S_2 = 1-2S_2 = 2 & \small \blue{\implies S_2 = -\frac 12} \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = 2+\frac 12 + 3S_3 = 3 & \small \blue{\implies S_3 = \frac 16} \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 = 3+1 + \frac 16 = \frac {25}6 \\ P_5 & = S_1P_4 - S_2P_3 + S_3P_2 = \frac {25}6+\frac 32 + \frac 13 = \boxed 6 \end{aligned}$

From the given equations we get $xy+yz+zx=-\dfrac{1}{2}$ , $xyz=\dfrac{1}{6}$ . So $x, y, z$ are the roots of the equation $X^3-X^2-\dfrac{1}{2}X-\dfrac{1}{6}=0$ . Therefore $x^4+y^4+z^4=x^3+y^3+z^3+\dfrac{1}{2}(x^2+y^2+z^2)+\dfrac{1}{6}(x+y+z)=3+1+\dfrac{1}{6}=\dfrac{25}{6}$ and $x^5+y^5+z^5=x^4+y^4+z^4+\dfrac{1}{2}(x^3+y^3+z^3)+\dfrac{1}{6}(x^2+y^2+z^2)=\dfrac{25}{6}+\dfrac{3}{2}+\dfrac{2}{6}=\dfrac{25+9+2}{6}=\boxed 6$