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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 problem, it seems that it will be useful to use the difference of cubes formula and/or difference of squares formula as well as some substitutions. I wasn't able to determine a useful method of solving this, however. Using a computer program, I was able to deduce that the desired values $(a,b,c)$ are $(2,20,25)$.

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Let $x,y,z\in\mathbb{R}$ such that $x^3 = 5, y^3 = 4$, and $z^3 = 2$. Then $y=z^2, y^2 = 2z$, and $yz = 2.$ We want to simplify $3\sqrt{x-y}$. Note that

$$\begin{align} (x-y)(x+y)^2&=(x-y)(x^2+2xy+y^2)\\ &=x^3-y^3+x^2y-xy^2\\ &=1+(xz)^2-2xz\\ &=(xz-1)^2. \end{align}$$

Hence

$$\begin{align} 3\sqrt{x-y} &= 3\cdot\dfrac{(xz-1)(x^2-xy+y^2)}{(x+y)(x^2-xy+y^2)}\\ &=3\cdot\dfrac{x^3z-x^2-x^2(yz)+xy+x(y^2)z-y^2}{x^3+y^3}\\ &=\dfrac{1}{3}(3z-3x^2+3xz^2)=z+xz^2-x^2\\ &=\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25}\\ &\Rightarrow (a,b,c) = (2,20,25). \end{align}$$

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Though @Simon Fraser has given a good answer, This nested radical dates back to Ramanujan and other similar ones like, $\sqrt{\sqrt[3]{28} - \sqrt[3]{27}}$ . There is also an identity for radicals of the form, $\sqrt{\sqrt[^3]{a}-\sqrt[^3]{b}}$ by Ramanujan himself, which is; (for $m,n$ arbitary)

$$\sqrt{m \sqrt[^3]{4m-8n}+n\sqrt[^3]{4m+n}}\\=\pm\frac{1}{3}\left(\sqrt[^3]{(4m+n)^2}+\sqrt[^3]{4(m-2n)(4m+n)}-\sqrt[^3]{2(m-2n)^2}\right)$$

A very good paper which focuses on denesting of these kind of radicals will be, this. The lecture also focuses on which $a$ & $b$'s are denestable (Pg. 05) and a short proof of the claim. I'll state a brief of it;

$$ \small{\text{Let a, b ∈ Q such that a/b is not a perfect cube in Q. Then,} \\\sqrt{\sqrt[^3]{a}+\sqrt[^3]{b}} \text{ can be denested if and only if there are integers } m,n \text{ such that } \\\frac{a}{b} = \frac{(4m−8n)m^3}{(4m+n)n^3}}$$

We can get the denested identity for $\sqrt{\sqrt[3]{5} - \sqrt[3]{4}}$ by letting $(m,n)=(1,1)$

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    $\begingroup$ Very nice answer with general formula. +1 $\endgroup$ – Paramanand Singh Dec 31 '19 at 12:43

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