Yeah sure.. Newton's Sums..

Let $P(x)=m^3+bm^2+cm+d$ with roots $x$ , $y$ and $z$ ; and let $S_k=x^k+y^k+z^k$ . We are given $S_1=0$ , $S_3=18$ and $S_7=2058$ . Now, by Newton's Sums we have:

$S_1=0=-b \Longrightarrow b=0$

$S_2=-bS_1-2c \Longrightarrow S_2=-2c \Longrightarrow c=-\dfrac{S_2}{2}$

$S_3=18=-bS_2-cS_1-3d \Longrightarrow d=-6$

By Vieta's formulas, we know that $xyz=-d$ , hence $xyz=6$ . This value is rational, but there are six possible permutations of the triple $(x,y,z)$ , hence the asked value is $6xyz=\boxed{36}$ .

To prove that the triple (x,y,z) is rational we can obtain $S_7$ by recursion:

$S_7=42c^2 \Longrightarrow 2058=42c^2 \Longrightarrow c=\pm 7$ .

So, there are two possible polynomials for $P(m)=m^3 \pm 7 m -6$ . $m^3-7m-6$ factors as $(m+1)(m+2)(m-3)$ showing that $(x,y,z)=(-1,-2,3)$ and all its permutations are rational.

You can definitely approach the problem using Newton's Sums (read the solution by @Alan Enrique Ontiveros Salazar) but we will approach the following problem in a different way by setting up some recurrence relations (although it's a bit long)

So, let $S_n = x^n+y^n+z^n$

We already know the following values of $S_n$ -

$S_0=x^0+y^0+z^0=3$

$S_1=x+y+z=0$

$S_3=x^3+y^3+z^3=18$

$S_7=x^7+y^7+z^7=2058$

Now, we would try to find out the values of $xyz$ and $xy+yz+zx$ . (You'll see further that why are we doing so)

Since $x+y+z=0$

$3xyz=x^3+y^3+z^3 \Rightarrow xyz=18/3 = 6$

Now, we just need to show that $(x,y,z)$ are rational. So,

$xy+yx+zx=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2} = -\frac{S_2}{2}$

Now, we would set up the recurrence relation for $S_n$ . We do so by writing up the characteristic equation for $S$ , which is -

$(S-x)(S-y)(S-z)=0$ $\Rightarrow S^3-(x+y+z)S^2+(xy+yz+zx)S-xyz=0$ $\Rightarrow S^3-\frac{S_2}{2}S-6=0$

[You can read about the recurrence relations here - http://www.webpages.uidaho.edu/~markn/395/pdf/rec-eq.pdf

And for more practice on recurrence equations, you may try out these problems by @Aditya Raut- https://brilliant.org/profile/aditya-5j3pzo/sets/bashing-unavailable-awesome-problems/]

Therefore, the recurrence relation that we generate from the above equation is -

$S_{n+3}= \frac{S_2S_{n+1}}{2}+6S_n$

Therefore, we get -

$S_5=\frac{S_2S_3}{2}+6S_2=9S_2+6S_2=15S_2$

$S_7=\frac{S_2S_5}{2}+6S_4=\frac{15}{2}S_2^2+6(\frac{S_2^2}{2}+6S_1)=\frac{21}{2}S_2^2$

Now observe that, $\frac{S_2}{2}\frac{S_3}{3}=\frac{S_5}{5}$

Similarly, we can also see that, $\frac{S_2}{2}\frac{S_5}{5}=\frac{S_7}{7}$

[You may be interested in solving this generalized version of the problem in USAMO 1982, Problem 2 - http://www.artofproblemsolving.com/Wiki/index.php/1982 USAMO Problems/Problem_2]

Therefore, after plugging in values of $S_3$ and $S_5$ , we get - $S_2=x^2+y^2+z^2=14$

This yields, $xy+yz+zx=-7$

Now, let $x,y,z$ be the roots of a cubic equation. This yields - $t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = t^3-7t-6 = (t+1)(t+2)(t-3)$

Hence, we find the rational solutions for $(x,y,z)=(-1,-2,3)$

Therefore, $\sum_{i=1}^n x_iy_iz_i$ is equal to the all the permutations of $(x,y,z)$ which yields $\sum_{i=1}^n x_iy_iz_i = 6+6+6+6+6+6=\boxed{36}$