Vietnamese Mathematical Olympiad

From the given system, it follows that
$$(6x^2 - 3y^2 - z - 2z^2) - 3(x^2 - y^2 - 2z + 1) - (6x - y + z^2 - 3) = 0.$$
This equation can be rewritten as $(x-z)(x+z-2) = 0$. Hence, $x = z$ or $x+z = 2$.

* If $x = z$. The given system is equivalent to
$$\begin{cases} x^2 + 6x - y = 3, \\ x^2 - 2x - y^2 = -1. \end{cases}$$
From the second equation, it obtains that $y^2 = (x-1)^2$ or $y = \pm (x-1)$. Put $y = 1-x$ and $1+x$ into the first equation, it obtains there are 4 solutions $(x, y, z)$ in this case, which are
$$\begin{pmatrix} \frac{-5 \pm \sqrt{33}}{2}, & \frac{-7 \pm \sqrt{33}}{2}, & \frac{-5 \pm \sqrt{33}}{2} \\ \frac{-7 \pm \sqrt{65}}{2}, & \frac{9 \mp \sqrt{65}}{2}, & \frac{-7 \pm \sqrt{65}}{2} \end{pmatrix}.$$
* If $x+z=2$. Letting $z=2-x$, the given system is equivalent to the following
$$\begin{cases} x^2 + 2x + 1 - y = 0, \\ x^2 + 2x - y^2 - 3 = 0. \end{cases}$$
It follows that $1 - y = -y^2 - 3$ or $y^2 - y + 4 = 0$. But this equation has no real roots, hence there is no solution for the given system in this case.
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