Let the ordered triples $(x,y,z)$ of complex numbers that satisfy

Question

\begin{align*}
x + yz &= 7, \\
y + xz &= -3, \\
z + xy &= -5.
\end{align*}
be $(x_1,y_1,z_1),$ $(x_2,y_2,z_2),$ $\dots,$ $(x_n,y_n,z_n).$ Find $x_1 + x_2 + \dots + x_n.$

Answer 1

If we add up all three equations, we get
\[x + yz + y + xz + z + xy = -1,\]which simplifies to $x + y + z + xy + xz + yz = -1.$ Then $(x + 1)(y + 1)(z + 1) = -2.$

By symmetry, $(z + 1)(x + 1)(y + 1) = -2.$ Subtracting these equations, we get
\[(x + 1)(z - y) = 0.\]If $x + 1 = 0,$ then $z + 1 = 0,$ so $y + 1 = 1,$ $y = 0,$ and $z = -1.$

Otherwise, $y = z,$ so
\[y + x^2 = 7\]and $y^2 + yx = -5.$ Subtracting these equations, we get
\[(x - y)(x + y + 1) = 12.\]From the equation $x + y^2 = 7,$ $(x + y) (x - y) = 7 - y^3 = 12.$ Hence,
\[y^3 = -5,\]so $y = \sqrt[3]{5} \omega,$ where $\omega$ is a cube root of unity.

(Note: The factorization above only works because we established that $y = z.$)

Then $(x + y)(x - y) = 12.$ Since $x + y = y + z$ is the sum of the roots, and $x - y = z - y$ is the product of the roots, by Vieta's formulas, the sum of all possible values of $x$ is $\boxed{3}.$