$x, y, z$ Polynomials

Relevant wiki: Algebraic Identities

$\begin{cases} \color{#3D99F6}{x + y + z} = 0 \ \\ \color{#20A900}{x^3 + y^3 + z^3}=18 \ \\ x^7 + y^7 + z^7=2058 \end{cases}$

$\Rightarrow \color{#20A900}{x^3+y^3+z^3}=18$

Adding $(\color{#D61F06}{-3xyz})$ to both sides.

$\color{#20A900}{x^3+y^3+z^3}\color{#D61F06}{-3xyz}=18\color{#D61F06}{-3xyz}$

$\underbrace{(\color{#3D99F6}{x+y+z})}_{0}(x^2+y^2+z^2-xy-yz-zx)=18\color{#D61F06}{-3xyz}$

$3xyz=18$
$\therefore xyz=\color{#BA33D6}{\boxed{6}}$

If $x+y+z = 0$ then $x^3+y^3+z^3 = 3xyz$ , because $x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) + 3xyz = 0 + 3xyz$ . Therefore, $xyz = \dfrac {18}3 = \boxed{6}$ .

If x+y+z=0 Then x=-(y+z).

x^3+y^3+z^3 = -(y+z)^3+y^3+z^3 = -3zy^2 - 3yz^2 = 18. Therefore zy^2 + yz^2 = -6

yz(y+z)=-6

xyz = 6.

Consider a polynomial $P(w)=(w-x)(w-y)(w-z)$ . We can see that $x,~y,~z$ are roots of $P(w)$ . Expanding it:

$P(w)=w^3-(x+y+z)w^2+(xy+xz+yz)w-xyz$

By Newton's Identities:

$a_3S(3)+a_2S(2)+a_1S(1)+a_0S(0)=0\\\\$

$1\cdot(x^3+y^3+z^3)-(x+y+z)(x^2+y^2+z^2)+(xy+xz+yz)(x^1+y^1+z^1)-xyz(x^0+y^0+z^0)=0\\\\$

$1\cdot18-0\cdot(x^2+y^2+z^2)+(xy+xz+yz)\cdot0-xyz\cdot3=0\\\\ 3xyz=18\\\\ \boxed{xyz=6}$