<s>[INST] How do I solve for $m$ and $n$ While reading about nested radicals, I came across a theorem that said $\sqrt{m\sqrt[3]{4m-8n}+n\sqrt[3]{4m+n}}=\pm\frac {1}{3}\left(\sqrt[3]{(4m+n)^{2}}+\sqrt[3]{4(4m+n)(n-2n)}+\sqrt[3]{(n-2n)^{2}}\right)$
So I tried an easy example ($\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}$) and got this System of Equation: $$\begin{cases}m^3(4m-8n)=-4\\n^3(4m+n)=5\end{cases}\text{or} \begin{cases}m^3(4m-8n)=5\\n^3(4m+n)=-4\end{cases}$$
My question: How do I solve for $m$ and $n$ without too much tedious work, and how do I know which System to discard?
 [/INST] Eliminate $m$ from the first system with
$$m=\frac{5-n^4}{4n^3},$$
giving
$$(5-n^4)^3(20-36n^4)=1024n^{12},$$
a quartic equation in $n^4$
$$9(n^4)^4-396(n^4)^3+750(n^4)^2-1500(n^4)+625=0.$$

Even Wolfram cannot find a closed-form solution !

With $-4$ in the RHS of the first equation, the equation is
$$9(n^4)^4+116(n^4)^3+750(n^4)^2-1500(n^4)+625=0,$$
which has the root $n^4=1$.
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