An interesting property of nested radicals I found a beautiful form of the sum of square roots.
$a,b,c\in\mathbb{R}$
\begin{eqnarray*}
    A &:=& abc\\
    B &:=& a+b+c\\
    C &:=& ab+bc+ca
\end{eqnarray*}
$$\sqrt{a}+\sqrt{b}+\sqrt{c} = \sqrt{A^0B+2\sqrt{A^0C+2\sqrt{A^1B+2\sqrt{A^2C+2\sqrt{A^5B+2\sqrt{A^{10}C}}}}}}\ \ \cdots\cdots$$
This property is based on the next relation.
$$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2 = a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$
$$(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2 = ab+bc+ca+2(\sqrt{ab^2c}+\sqrt{abc^2}+\sqrt{a^2bc})$$
$$(\sqrt{ab^2c}+\sqrt{abc^2}+\sqrt{a^2bc})^2 = ab^2c+abc^2+a^2bc+2(\sqrt{a^2b^3c^3}+\sqrt{a^3b^2c^3}+\sqrt{a^3b^3c^2})$$
$$\vdots$$
In this way, three sums of square roots generate three sums of square roots. Therefore, it's a recursive structure.
$$$$
The same property can be confirmed for the cube roots.
$a,b\in\mathbb{R}$
\begin{eqnarray*}
    A &:=& ab\\
    B &:=& a+b
\end{eqnarray*}
$$\sqrt[3]{a}+\sqrt[3]{b} = \sqrt[3]{A^0B+3\sqrt[3]{A^1B+3\sqrt[3]{A^4B+3\sqrt[3]{A^{13}B+3\sqrt[3]{A^{40}B+3\sqrt[3]{A^{121}B}}}}}}\ \ \cdots\cdots$$
Please let me know if you have any interesting information related to these.

I don't know why, but it seems that the exponent of A in the case of the cube root is in this sequenceA003462.
