How do I solve for $m$ and $n$

Question

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?

Answer 1

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 $n^4$

$$9(n^4)^4-396(n^4)^3+750(n^4)^2-1500(n^4)+625=0.$$

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Even Wolfram cannot find a closed-form solution !

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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 $n^4=1$.

Answer 2

$$m^{3}(4m-8n)=4=m^4(4-8\frac{n}{m})$$ similarly $$n^4(4\frac{m}{n}+1)=5$$ now let $\frac{m}{n}=t$ ,therefore $$m^4(4-\frac{8}{t})=4...(1)$$ and $$n^4(4t+1)=5...(2)$$
now divide both equations $(1) \& (2)$ to have a equation solely in $t$ and after solving for $t$ you can get $m$ and $n$. as far as solution to choose , choose whichever gives you a postive answer as square root cannot be negative (p.s this is tedious but the only approach i could think of)