Forming quadratic

Let $u=-z$ . Then $x+y+u=0$ and

$\begin{aligned} x^3+y^3 + u^3 & = - 3 \\ (\blue{x+y+u})(x^2+y^2+u^2 - xy-yu-ux) + 3\red{xy}u & = - 3 & \small \blue{\text{Note that }x+y+u = 0} \\ \blue 0 + 3 \cdot \red 1 \cdot u & = - 3 & \small \red{\text{and }xy=1} \\ \implies u & = -1 \end{aligned}$

From $x+y+\blue u=0$ multiplied by $x$ on both sides, we have:

$\begin{aligned} x^2 + \red{xy} + \blue{(-1)}x & = 0 \\ x^2 - x + 1 & = \boxed 0 \end{aligned}$

Cubing both side the equation (1), we will obtain $x^3 + y^3 + 3xy(x+y) = z^3 --- (4)$ Thus, by (1) and (2), equation (4) imply that $(z^3 -3) + 3xyz = z^3 --- (5)$ By equation (3), the equation (5) imply that $-3 + 3z = 0$ , and hence, $z= 1$ . Thus, equation (1) will become $x+y = 1$ . Observe that $x$ and $y$ are roots of the quadratic equation $a^2 -a +1 = 0$ with sum of the roots $x+y=1$ and product of the roots $xy = 1$ . Therefore, $x^2 -x +1 = 0$ as desired.