This is the last problem of Newton's Sums I'll post

Let $S_1=x+y+z$ , $S_2=xy+xz+yz$ , $S_3=xyz$ and $P_n=x^n+y^n+z^n$ .

By Newton's Sums we have:

$P_1=S_1$

$P_2=S_1^2-2S_2$

$P_3=S_1^3-3S_1S_2+3S_3$

$P_4=S_1^4-4S_1^2S_2+4S_1S_3+2S_2^2$

$P_5=S_1^5 -5S_1^3S_2 +5S_1^2S_3 +5S_1S_2^2 -5S_2S_3$

First notice that the value asked is $P_2+P_5$ .

With the known values the following system is formed:

$\quad \begin{aligned} 3=S_1 &\quad ...(1) \\15=S_1^3-3S_1S_2+3S_3 &\quad ...(2) \\ 35=S_1^4-4S_1^2S_2+4S_1S_3+2S_2^2 &\quad ...(3) \end{aligned}$

From $(1)$ it's easy to see that $S_1=3$ . Substitute it in $(2)$ and $(3)$ :

$\quad \begin{aligned} 15=27-9S_2+3S_3 &\quad ...(4) \\35=81-36S_2+12S_3+2S_2^2 &\quad ...(5) \end{aligned}$

Solve for $S_3$ in $(4)$ and substitute in $(5)$ :

$S_3=3S_2-4$

$35=81-36S_2+12(3S_2-4)+2S_2^2$

$S_2^2=1 \Longrightarrow S_2=\pm1$

First let's work with $S_2=1$ . Obtain $S_3$ :

$S_3=3(1)-4=-1$

Now obtain $P_2$ and $P_5$ :

$P_2=(3)^2-2(1)=7$

$P_5=(3)^5-5(3)^3(1)+5(3)^2(-1)+5(3)(1)^2-5(1)(-1)=83$

So, in this case, $P_2+P_5=\boxed{90}$ .

Now, let's work with $S_2=-1$ . Obtain $S_3$ :

$S_3=3(-1)-1=-7$

Now obtain $P_2$ and $P_5$ :

$P_2=(3)^2-2(-1)=11$

$P_5=(3)^5-5(3)^3(-1)+5(3)^2(-7)+5(3)(-1)^2-5(-1)(-7)=43$

So, in this case, $P_2+P_5=\boxed{54}$ .

FInally, the sum of the two possible values of $P_2+P_5$ is: $90+54=\boxed{144}$ .