A constant bizarrely related to the Fibonacci Numbers 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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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, 
$\frac{3}{2}(\frac{\sqrt[3]{2}}{\sqrt[3]{2}+1})^3=\sqrt[3]{2}-1$
