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An interesting property of nested radicals

I found a beautiful form of the sum of square roots.

$a,b,cinmathbb{R}$
begin{eqnarray*}
A &:=& abc\
B &:=& a+b+c\
C &:=& ab+bc+ca
end{eqnarray*}

$$sqrt{a}+sqrt{b}+sqrt{c} = sqrt{A^0B+2sqrt{A^0C+2sqrt{A^1B+2sqrt{A^2C+2sqrt{A^5B+2sqrt{A^{10}C}}}}}} cdotscdots$$

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.

$a,binmathbb{R}$
begin{eqnarray*}
A &:=& ab\
B &:=& a+b
end{eqnarray*}

$$sqrt[3]{a}+sqrt[3]{b} = sqrt[3]{A^0B+3sqrt[3]{A^1B+3sqrt[3]{A^4B+3sqrt[3]{A^{13}B+3sqrt[3]{A^{40}B+3sqrt[3]{A^{121}B}}}}}} cdotscdots$$

Please let me know if you have any interesting information related to these.

Mathematics Asked by isato on November 21, 2021

1 Answers

One Answer

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

Answered by ratbase on November 21, 2021

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