Roots of a set of nonlinear equations $ax + yz = b_1; ay + xz = b_2; az + xy = b_3$ Let $a$ be a non-negative real number, $ b_1, b_2, b_3$ be real numbers, and $x, y, z$ be variables. Is it possible to analytically find the roots of the set of nonlinear equations given by
$$\begin{cases}
ax + yz =  b_1\\
ay + xz =  b_2\\
az + xy =  b_3\;\;\; ?\end{cases}$$
 A: Possible Strategy.
We add the $1$st and $2$nd line equations side-by-side and then subtract:
$\begin{align}&(x+y)(a+z)=b_1+b_2\\
&(x-y)(a-z)=b_1-b_2\end{align}$.
This implies that,
$\begin{align}&x=\frac 12\left(\frac {b_1+b_2}{a+z}+\frac {b_1-b_2}{a-z}\right)\\&y=\frac 12\left(\frac {b_1+b_2}{a+z}-\frac {b_1-b_2}{a-z}\right)\end{align}$.
Finally, we use this result in the $3$rd line equation $az+xy=b_3\,:$
$\begin{align}az+\frac 14\left(\frac {(b_1+b_2)^2}{(a+z)^2}-\frac {(b_1-b_2)^2}{(a-z)^2}\right)-b_3=0.\end{align}$
Now, you have an polynomial equation depends on a single variable.  We can continue from here.
$\begin{align}4az\left(a^2-z^2\right)^2+(a-z)^2(b_1+b_2)^2-(a+z)^2(b_1-b_2)^2-4b_3\left(a^2-z^2\right)^2=0\end{align}.$
Since the degree of the polynomial we get is $5$, hence we will need knowledge of Galois theory.
Depending on how we choose the coefficients, a solution with radicals may be possible.  However, since we are looking for an exact solution, it is possible to say that the polynomial cannot be solved by radicals, in general.
