Solving equation in three variables please help me understand how the following equation with 3 variables and power of 2 is solved and what solution approach is the quickest.  
$$3y^2 - 3 = 0$$
$$4x - 3z^2 = 0$$
$$-6xz+ 6z = 0 $$
 A: $$
\begin{cases}
3y^2-3=0\\
4x-3z^2=0\\
6z-6xz=0
\end{cases}\Longleftrightarrow
\begin{cases}
3y^2-3=0\\
x=\frac{3z^2}{4}\\
6z(1-x)=0
\end{cases}\Longleftrightarrow
\begin{cases}
3y^2-3=0\\
x=\frac{3z^2}{4}\\
6z\left(1-\frac{3z^2}{4}\right)=0
\end{cases}
$$
So, when you solve:
$$6z\left(1-\frac{3z^2}{4}\right)=0$$
We get three solutions for $z$, $z=0$ or $z=\pm\frac{2}{\sqrt{3}}$
A: A quick solution:
By simplification, rewrite
$$\begin{cases}y^2=1\\(1-x)z=0\\3z^2=4x.\end{cases}$$
Then mentally,
$$y=\pm1,\\x=z=0\lor x=1,z=\pm\frac2{\sqrt3}.$$
A: $3y^2-3=0.............(1)$
$4x-3z^2=0...........(2)$
$6z-6xz=0...........(3)$
First, solving the equation(1),
$3y^2-3=0\;\;\;\implies y^2=1\;\;\;\implies y=\pm1$
Now, solving the equation(3),
$6z-6xz=0\;\;\;\implies 6z(1-x)=0$
$z=0,\;1-x=0$
$z=0,\;x=1$
Solving the equation(2),
$4x-3z^2=0$
Put the value of $x$ in equation(2)
$4(1)-3z^2=0\;\;\implies-3z^2=-4\;\;\implies z=\sqrt{\dfrac{4}{3}}\;\implies z=\pm \dfrac{2}{\sqrt{3}}$
Thus, $x=1,\;y=\pm1,\;z=0\;or\;z=\pm \dfrac{2}{\sqrt{3}}$
