Question: Given $$\begin{cases}m=u^5-40u^2v^3+96v^5\\n=6u^5-64v^5\end{cases}\tag{1}$$ Find $u,v$ in terms of $m,n$.
If that was a bit confusing, I have provided a different example below:
Example:$$m=u^3+8v^3\\n=-4u^3+4v^3\tag{2}$$ Adding $n+4m$ gets rid of the $u$ term, and similarly for $m-2n$. Thus,$$n+4m=-4u^3+4v^3+4u^3+32v^3=36v^3\\-2n+m=8u^3-8v^3+u^3+8v^3=9u^3\\\therefore u=\left[(m-2n)/9\right]^{1/3}\\v=\left[(n+4m)/36\right]^{1/3}$$
I don't know what to do. (I have Mathematica, in case things get messy or I get desperate)
I'm thinking of substituting $u$ or $v$ with something, and hoping to get an expression, where I can substitute the variable back to eliminate it, but even that I'm not too sure where to begin.
The following outlines a brute force approach which reduces the problem to solving a quintic. Let $a=u^5, b=u^2 v^3,c=v^5$ then solve the system for $a,c$ with $b$ as parameter:
$$ \begin{cases} a + 96 c = m + 40 b \\ 6 a - 64 c = n \end{cases} $$
As expected, the solutions are linear in $b$ (though not particularly pretty):
$$ \begin{cases} a = \frac{1}{20} (80 b+2 m+3 n) \\ c = \frac{1}{640} (240 b+6 m-n) \tag{1} \end{cases} $$
Substituting the expressions above into the identity $b^5 = a^2 c^3$ gives a quintic in $b$. If that quintic happens to be solvable by radicals, then it would give $b$ as a function of $m,n$, which substituted back into $(1)$ would give $a,c$ thus $u,v$. I don't see an obvious nice closed form, though.