Solving a system of quadratic equations. Solve for real $(a, b, c)$ satisfying
$$ab + bc + ca = 1$$
$$a^2 − 2b^2 = 1$$
$$2b^2 − 3c^2 = 1$$
I try isolating $a$, but it leads to a very complicated expression in $a$.
 A: If one can show that $c=0$ then $a=\sqrt{2},b=\sqrt{2}/2$ as noted in Tim's answer. I found it a bit involved to actually prove $c=0$ is a consequence of the equations. (I'd be interested in a simpler proof of $c=0$ than the following.) Label the equations:
[1] $ab+bc+ca=1,$
[2] $a^2-2b^2=1,$
[3] $2b^2-3c^2=1.$
From [2] and [3] we can see neither of $a,b$ are zero, since we seek real solutions. From [2] alone we can also see that neither of $a+b,a-b$ are zero, otherwise [2] says $-t^2=1$. We'll need these nonzero properties later.
Now since the right sides are all equal, we have from [1] and [2] that
$$c(a+b)+ab=a^2-2b^2,$$
and moving the $ab$ to the right side and factoring gives
$$c(a+b)=(a+b)(a-2b).$$
As noted we know $a+b$ is nonzero, so we now have $c$ in terms of $a,b$ as
$$c=a-2b.$$
We next use that [2] and [3] have the same right sides, so that on replacing $c$ as above,
$$a^2-2b^2=2b^2-3(a-2b)^2,$$ and the difference factors to obtain:
$$4(a-2b)(a-b)=0.$$
Having noted that we know $a-b$ is not zero we arrive at $a-2b=0$, i.e. $c=0$ as desired.
EDIT: I removed a phrase "luckily the $b^2$ terms cancel" since they didn't.
Also as others have noted, in case $c=0$ the signs on $a,b$ may be changed for another solution, so that the system has actually two triples $(a,b,c)$ for solutions.
A: As one approach try putting the first equation $ab+bc+ca=1$ in terms of $c$ then set $c$ equal to $0$. It will look kind of complicated as you put it with variable terms under radical signs but you will see that the result equals $1$. From there the other two solutions for $a$ and $b$ will be apparent. Real solutions are $$a=\sqrt{2}$$ $$b=\frac{\sqrt{2}}{2}$$ $$c=0$$ to check your work.$$ab=\frac{\sqrt{18c^4 +18c^2+4}}{2}$$
$$bc=\frac{c\sqrt{2+6c^2}}{2}$$
$$ca=\frac{2c\sqrt{2+3c^2}}{2}$$
