<s>[INST] Triples $(x, y, z)$ that satisfy a set of equations 
Suppose that $a$ is a  fixed (but unknown) real number, with $a^2 \neq 1$.  Determine all triples $(x, y, z)$ of real numbers that satisfy the system of equations:
$x + y + z = a$
$xy + yz + xz = -1$
$xyz = -a$

I've tried making substitutions but don't seem to be able to make much progress. Another thing I noticed is that:
$(x + y + z) = a \implies$
$(x + y + z)^2 = a^2 \implies$
$x^2 + y^2 + z^2 + 2(xy + yx + xz) = a^2 \implies$
$x^2 + y^2 + z^2 + 2(-1) = a^2 \implies$
$x^2 + y^2 + z^2 = a^2 + 2$
but I don't know if that actually is useful. Any suggestions on how to move forward?
 [/INST] We have
\begin{align}
(u-x)(u-y)(u-z) &= u^3 - (x+y+z)u^2 + (xy+yz+xz)u - xyz\\
&= u^3-au^2-u+a\\
&= u^2(u-a)-(u-a)\\
&= (u^2-1)(u-a)\\
&= (u-1)(u+1)(u-a)
\end{align}
so $\{x,y,z\} = \{-1,1,a\}$.
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