A Summation?

Let $P_n = x^n+y^n+z^n$ , where $n$ is a positive integer, and $S_1 = x+y+z=1$ , $S_2 = xy +yz+zx$ , and $S_3 = xyz$ . Then by Newton's sums or identities , we have:

$\begin{aligned} P_1 & = S_1 = 1 \\ P_2 & = S_1 P_1 - 2S_2 = 2 & \small \blue{\implies S_2 = - \frac 12} \\ P_3 & = S_1 P_2 - S_2 P_1 + 3S_3 = 2 + \frac 12 + 3S_3 = 3 & \small \blue{\implies S_3 = \frac 16} \\ P_4 & = S_1 P_3 - S_2 P_2 + S_3P_1 = 3 + 1 + \frac 16 = \frac {25}6 \\ P_5 & = S_1 P_4 - S_2 P_3 + S_3P_2 = \frac {25}6 + \frac 32 + \frac 13 = \boxed 6 \end{aligned}$

From the first two equations we get

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

$xy+yz+zx=-\dfrac{1}{2}$ .

Using the third equation we get

$xyz=\dfrac{1}{3}[x^3+y^3+z^3-(x+y+z)(x^2+y^2+z^2-xy-yz-zx)]=\dfrac{1}{3}[3-1\times {(2+\dfrac{1}{2})}=\dfrac{1}{6}$ .

Hence $x, y, z$ are the roots of the equation $X^3-X^2-\dfrac{1}{2}X-\dfrac{1}{6}=0$ . Hence

$x^4+y^4+z^4=x^3+y^3+z^3+\dfrac{1}{2}(x^2+y^2+z^2)+\dfrac{1}{6}(x+y+z)=3+1+\dfrac{1}{6}=\dfrac{25}{6}$ ,

and

$x^5+y^5+z^5=x^4+y^4+z^4+\dfrac{1}{2}(x^3+y^3+z^3)+\dfrac{1}{6}(x^2+y^2+z^2)=\dfrac{25}{6}+\dfrac{3}{2}+\dfrac{1}{3}=\boxed 6$