So I came across this question the other day asking me to solve the following system of equations:
$$ \left\{ \begin{array}{c} x^2-yz=1 \\ y^2-zx=2 \\ z^2-xy=3 \end{array} \right. $$
Normally when solving these 3 variable systems of equations I would use the method where you put the coefficients into a matrix. However, to me it seems like that wouldn't work here, unless there is some trick that needs to happen. Are there any easy-ish ways of solving this? Or do I just need to use good old substitution until I can isolate some variables? Substitution is what I tried originally, it just seemed to get really long and I gave up.
HINT
The proposed system of equations can be rewritten as \begin{align*} \begin{cases} x^{2} - yz = 1\\\\ y^{2} - zx = 2\\\\ z^{2} - xy = 3 \end{cases} & \Longleftrightarrow \begin{cases} x^{2} - yz = 1\\\\ y^{2} - x^{2} + yz - zx = 1\\\\ z^{2} - y^{2} + zx - xy = 1 \end{cases} \Longleftrightarrow \begin{cases} x^{2} - yz = 1\\\\ (y-x)(x + y +z) = 1\\\\ (z-y)(x + y + z) = 1 \end{cases}\\\\ & \Longleftrightarrow \begin{cases} x^{2} - yz = 1\\\\ (y-x)(x+y+z) = 1\\\\ (z-y)(x+y+z) = 1\\\\\ (z - 2y + x)(x+y+z) = 0 \end{cases} \end{align*}
The last equation tells us that $x = 2y -z$ or $x = -y - z$. But $x+y+z$ cannot be equal to zero (why?). Thus it remains just the first option.
Can you take it from here?