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?
 A: Make a polynomial $$P(t) =t^3 -at^2 -t+a$$
By Vieta formulas, $x,y,z$ are roots of $P(t)$ and it is easy to calculate them:
$$ P(t) = t(t^2-1)-a(t^2-1)= (t-1)(t+1)(t-a)$$
so $x,y,z$ are permutation of $1,-1,a$.
A: 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 $\{x,y,z\} = \{-1,1,a\}$.
A: A more systematic way: Use the third equation to eliminate $z$ in the second equation, 
$$xy-a/x -a/y = -1$$
$$z = -a/xy \Rightarrow x+y-\frac{a}{xy} = a$$
Now spot $-a/x - a/y = -a\frac{x+y}{xy}$. 
So the first equation can be written as $$a(x+y) = (xy)^{2}+(xy)$$
Getting rid of $x+y$, the second equation becomes
$$(xy)^{2}+(xy)-a^{2}/(xy) = a^{2}$$
Aha! This is a cubic equation in $w=xy$. Precisely, 
$$w^{3}+w^{2}-a^{2}w-a^{2} = w^{2}(w+1)-a^{2}(w+1) = (w^{2}-a^{2})(w+1)=0$$
We have three possibilities: $w=a$,$w=-a$ and $w=-1$. 
Note that $zw=-a$, so we can just consider the possibilities $z=1,-1,a$. If $z$ is known, then 
$$y=a-z-x \Rightarrow x(a-z-x) +z(a-z) = -1$$
which is a quadratic in $z$. The rest is enumeration, good luck!
