If \(x^{2} = y + z\), \(y^{2} = z + x\) and \(z^{2} = x + y\), where \(x + y + z \neq 0\), then the…

The value is 1. Adding x to both sides of $x^{2} = y + z$ gives $x(x + 1) = x + y + z$; since this sum is non-zero, $x + 1 \neq 0$ and $\frac{1}{x + 1}$ $= \frac{x}{x + y + z}$. In the same way $\frac{1}{y + 1}$ $= \frac{y}{x + y + z}$ and $\frac{1}{z + 1}$ $= \frac{z}{x + y + z}$. Adding the three gives $\frac{x + y + z}{x + y + z}$ = 1 for every such x, y and z, so −1, 2 and 4 are not possible.