I have a system of 3 equations with the unknown $(x,y,z)$ to solve in the space $\mathbb{R}^2 \times \mathbb{R}_+^{*}$ : $$m=x-y \\ v=x^2(z-1) \\\gamma=x^3(z^3+3z-2)$$
with $m \in \mathbb{R}$, $v>0$ , $\gamma \in \mathbb{R}$ and $z>1$
Numerically, I easily find the solution with a solver, but I cannot figure out whether a closed-form formula of the solution exists.
Thanks.
A standard way to solve the last two equations, would be to square the last and cube the other to solve for $z$ (expect when $x=0$, but in this case $y=-m$ and $z$ can be anything): $$ v³(z³+3z-2)²-\gamma^2(z-1)³ = 0 $$ This gives a sixth order polynomial that you still have to check for extraneous solutions (as we have squared and lost the sign of $\gamma$ and $x$). But than, using the second, we get: $$ x = \pm\sqrt{\frac{v}{z-1}} $$ Which also requires $z>1$ for a solution, and finally using the first one: $$ y = x - m $$