In Lawrence Washington's book on Elliptic Curves section 2.5.2 on Cubic Equations, starting from $x^3 + y^3 + z^3 = 0$ such that $xyz \neq 0$, he derives $$\frac{x}{z} = u + v, \qquad \frac{y}{z} = u - v$$ $$(u + v)^3 + 6uv^2 + 1 = 0 \implies 6(v/u)^2 = -(1/u)^3 - 2$$ Let $$x_1 = \frac{-6}{u} = -12 \frac{z}{x + y}, \qquad y_1 = \frac{36v}{u} = 36 \frac{x - y}{x + y}$$ Then $$y_1^2 = x_1^3 - 432$$ So far so good. But then he states $(x_1, y_1) = \infty$ is a solution, and that $(x_1, y_1) = \infty$ corresponds to $x = -y \implies z = 0$.
How did he get from $(x_1, y_1) = \infty \implies x = -y$? I can see how if we used projective coordinates, then the identity value would mean $z = 0$, but how did he first get $x = -y$? Where does this derivation come from?
Washington avoids projective coordinates.
In projective space, all affine solutions have the form $(x,y,1)$ with last coordinate 1, while the solutions at infinity are of shape $(x,y,0)$ with $x\ne 0\ne y$. Here $(x,-x,0)$ is a solution at infinity.