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.