Linked Questions

I have recently read some passage about nested radicals, I'm deeply impressed by them. Simple nested radicals $\sqrt{2+\sqrt{2}}$,$\sqrt{3-2\sqrt{2}}$ which the later can be denested into $1-\sqrt{2}$....

As an example, consider the polynomial $f(x) = x^3 + x - 2 = (x - 1)(x^2 + x + 2)$ which clearly has a root $x = 1$. But we can also find the roots using Cardano's method, which leads to $$x = \sqrt[...

Denest $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$. I have tried completing square by several method but all failed. Can anyone help me please? Thank you. p.s. I'm a poor question-tagger.

I have been looking into denesting square roots but I have found that $\sqrt[3]{2+\sqrt{5}}$ equals $(1+\sqrt{5})/2$. The same is true for $\sqrt[3]{2-\sqrt{5}}$ and $(1-\sqrt{5})/2$. I cannot figure ...

It is well known that $$\operatorname{R}(-e^{-\pi})=-\cfrac{e^{-\frac{\pi}{5}}}{1-\cfrac{e^{-\pi}}{1+\cfrac{e^{-2\pi}}{1-\cfrac{e^{-3\pi}}{1+\ddots}}}}=\frac{\sqrt{5}-1}{2}-\sqrt{\frac{5-\sqrt{5}}{2}}$...

Online, one may find the values of the following sums: $$\sum_{k=-\infty}^\infty e^{-\pi k^2}=\frac{\pi^{1/4}}{\Gamma(3/4)}$$ $$\sum_{k=-\infty}^\infty e^{-2\pi k^2}=\frac{\pi^{1/4}(6+4\sqrt 2)^{1/4}}{...

I've come across a very curious Ramanujan identity $$\sqrt[6]{7 \sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}$$ You could probably prove this by taking the 6th power of both ...

I have seen several questions asking for the proof of $$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$$ However, I want to simplify $$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}...

How check that $ \sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\sqrt[3]{\sqrt[3]2-1} $?

Is it possible to represent $$ \sqrt[3] {7\sqrt[3]{20}-1} =\sqrt[3]{A}+\sqrt[3]{B}+\sqrt[3]{C}$$ with rational $A,\,B,$ and $C?$

I'm wondering if there is a way to simplify the nested cubic radicals $$\sqrt[3]{\sqrt[3]{A}-B}$$ into its denested form $$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$$ Examples include $$\sqrt[3]{7\sqrt[3]{...

Let $a,b,c$ be positive integers such that $\sqrt{\sqrt[3]{5} - \sqrt[3]{4}}\times 3 = \sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c}$. Determine the values of $a, b,$ and $c$. To solve this simplification ...

I came across a problem on the web, it is as follows: Question:If $ x $, $ y $ and $ z $ are rational numbers such that $ \sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}$ then ...

$$\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$$ I could not multiply by the conjugate since it is a cube root. Can you show me a way to simplify it? Thanks!

Edit: My question comes from finding the solutions of this equation using Cardano's method(because our teacher said :D ): $$x^3-6x+4=0$$ And finally I got: $$x=(\sqrt[3]{2})\sqrt[3]{-2+\sqrt{-1}}⠀+(\...