<s>[INST] Strategies to denest nested radicals $\sqrt{a+b\sqrt{c}}$ 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 $1-\sqrt{2}$. This may be able to see through easily, but how can we denest such a complicated one $\sqrt{61-24\sqrt{5}}(=4-3\sqrt{5})$? And Is there any ways to judge if a radical in $\sqrt{a+b\sqrt{c}}$ form can be denested?
Mr. Srinivasa Ramanujan even suggested some CRAZY nested radicals such as:
$$\sqrt[3]{\sqrt{2}-1},\sqrt{\sqrt[3]{28}-\sqrt[3]{27}},\sqrt{\sqrt[3]{5}-\sqrt[3]{4}},
\sqrt[3]{\cos{\frac{2\pi}{7}}}+\sqrt[3]{\cos{\frac{4\pi}{7}}}+\sqrt[3]{\cos{\frac{8\pi}{7}}},\sqrt[6]{7\sqrt[3]{20}-19},...$$
Amazing, these all can be denested. I believe there must be some strategies to denest them, but I don't know how.
I'm a just a beginner, can anyone give me some ideas? Thank you.
 [/INST] There is also another radical identity, which generalises $\sqrt{a+\sqrt b}$ even further: \begin{align*}\sqrt{a\pm \sqrt{b+\sqrt{c}}}&= \frac{1}{\sqrt{2}}\left\{\sqrt{a^2+\sqrt{\frac{a^2-b+\sqrt{(a^2-b)^2-c}}{2}}-\sqrt{\frac{a^2-b-\sqrt{(a^2-b)^2-c}}{2}}}\right. \nonumber\\&\quad\left.\pm\,\sqrt{a^2-\sqrt{\frac{a^2-b+\sqrt{(a^2-b)^2-c}}{2}}+\sqrt{\frac{a^2-b-\sqrt{(a^2-b)^2-c}}{2}}}\right\}\end{align*} such that the $\pm$ signs are not independent of each other.
We also have:
If $a^2+b^2=c^2$, then $$(a+b-c)^2=2(c-a)(c-b).$$ Therefore, by letting $a=\sqrt x$ and $b=\sqrt{c^2-x}$, it follows for all $c$ and $x$ such that $c\in \mathbb{R}\setminus \big({-\sqrt x}, \sqrt x\big)$, $$\sqrt{2\big(c-\sqrt x\big)\big(c-\sqrt{c^2-x}\big)}=\sqrt x + \sqrt{c^2-x} -c.$$ Interestingly, given that $$(a+x)^2+(b+y)^2-(a+y)^2-(b+x)^2=2(a-b)(x-y)$$ then if $a^2+b^2=c^2$, we also have $$(a+b-c)^2=(2c)^2+(a+b)^2-(b+c)^2-(a+c)^2.$$ Through one of Ramanujan's classic identities, $$\big\{\sqrt [3]{(a+b)^2}-\sqrt [3]{a^2-ab+b^2}\big\}^3=3\big(\sqrt [3]{a^3+b^3}-a\big)\big(\sqrt [3]{a^3+b^3}-b\big)$$ it can be found that if $a^3+b^3=c^3$, then $$(a+b-c)^3=3(a+b)(c-a)(c-b)$$
Therefore, again, for all $c$ and $x$ such that $c\in\mathbb{R}\setminus ({-\sqrt [3] x}, \sqrt [3] x)$, it follows
$$\sqrt [3]{3\big(\sqrt [3] x+\sqrt [3]{c^3-x}\big)\big(c-\sqrt [3] x\big)\big(c-\sqrt [3]{c^3-x}\big)}=\sqrt [3] x + \sqrt [3]{c^3-x}-c.$$ Unfortunately, due to Fermat's Last Theorem, there are no positive integer solutions for unique $a$, $b$ and $c$.

Try such substitutions yourself on the following one! :)
If $a^4+b^4=c^4$, then $$(a+b-c)^4=6(ab-ac-bc)^2-4\big\{a^3(c-b)+b^3(c-a)+c^3(a+b)\big\}$$
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