For roughly the past month, I have been studying denesting radicals. For example: the expression $\sqrt[3]{\sqrt[3]2-1}$ is a radical expression that contains another radical expression, so this radical is *nested*. Is there a way to express this with radicals that are not (or not as) nested? Writing it in this way is referred to as *denesting*, and indeed, there is one such way. Ramanujan mysteriously found that $$\sqrt[3]{\sqrt[3]2-1}=\sqrt[3]{\frac 19}-\sqrt[3]{\frac 29}+\sqrt[3]{\frac 49}.$$ I found that this is also equal to $\sqrt{\sqrt[3]{\frac 43}-\sqrt[3]{\frac 13}}$, which does not denest it, but it does write the expression under a radical of a coprime degree, which fascinates me.


I have found an abundance of results, one of which fascinates me the most, and is on the constant, $$1-\sqrt[3]{\frac 12}+\sqrt[3]{\frac 14}.$$ This constant is equal to all of the expressions below. Note that the degree of the radicals are, lo and behold, the Fibonacci numbers! $$\sqrt{\frac 32\bigg(\sqrt[3]2-\frac 1{\sqrt[3]2}\bigg)}$$ $$\sqrt[3]{\frac{3^2}{2^2}\big(\sqrt[3]2-1\big)}$$ $$\sqrt[5]{\frac{3^3}{2^3}\bigg(\frac 3{\sqrt[3]2}-\sqrt[3]2-1\bigg)}$$ $$\sqrt[8]{\frac{3^5}{2^5}\bigg(4-\frac{5}{\sqrt[3]2}\bigg)}$$ $$\sqrt[13]{\frac{3^8}{3^8}\bigg(1+\frac{17}{\sqrt[3]2}-\frac{23}{\sqrt[3]4}\bigg)}$$ and, slightly breaking the pattern, $$\sqrt[21]{\frac{3^{14}}{2^{14}}\big(41-59\sqrt[3]2+21\sqrt[3]4\big)}$$ and presumably, this list goes on forever (but the numbers start becoming pretty big).


So... what on earth is going on here? It appears that for some $n$th Fibonacci number $F\_n$, this is equal to (for at least most of the time), $$\sqrt[F\_n]{\frac{3^{F\_{n-1}}}{2^{F\_{n-1}}}\big(a+b\sqrt[3]2+c\sqrt[3]4\big)}$$ for some $\{a, b, c\}\subset \mathbb R$, with a minimal polynomial of $4x^3-12x^2+18x-9$.


Can anybody explain these wild affairs?




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Don't know of any other appropriate tags



A constant bizarrely related to the Fibonacci Numbers
First of all we get $A=1-\sqrt[3]{\frac{1}{2}}+\sqrt[3]{\frac{1}{4}}=\frac{3\sqrt[3]{2}}{2(\sqrt[3]{2}+1)}$......(1)


And from here you can proof by induction...


$A^{F\_n}=A^{F\_{n-1}}.A^{F\_{n-2}}$


If $A^{F\_{n-1}}=(\frac{3}{2})^{F\_{n-2}}(a\_{n-1}+b\_{n-1}\sqrt[3]{2}+c\_{n-1}\sqrt[3]{4})$ 


and


$A^{F\_{n-2}}=(\frac{3}{2})^{F\_{n-3}}(a\_{n-2}+b\_{n-2}\sqrt[3]{2}+c\_{n-2}\sqrt[3]{4})$


From (1) we get, $A^{F\_2}, A^{F\_3}$, so we shall get the rest by induction.


Where, $F\_2=2, F\_3=3$. 


$\frac{3}{2}(\frac{\sqrt[3]{2}}{\sqrt[3]{2}+1})^2=\sqrt[3]{2}-\sqrt[3]{\frac{1}{2}}$


And, 


$\frac{3}{2}(\frac{\sqrt[3]{2}}{\sqrt[3]{2}+1})^3=\sqrt[3]{2}-1$


