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In the previous post I proved an identity in radicals:

$\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}}=3$

which followed the method used in another post where another identity in radicals

$\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} = 1.$

has been derived. The latter post pointed to an earlier and perhaps a more exciting one where we established an unexpected identity

$\displaystyle\sqrt[3]{2 \pm \sqrt{5}} = \frac{1 \pm \sqrt{5}}{2}.$

This was verified by simply taking the cube of the two sides of the equation. In a private correspondence related to the previous post Bruce Reznick suggested a different way of showing that

$\sqrt[3]{18+\sqrt{325}}=\sqrt[3]{18 \pm 5 \sqrt{13}}=a+b\sqrt{13},$

with rational $a$ and $b.$ In Bruce Reznick's words:

So we look for $(a + b \sqrt{13})^3 = 18 + 5 \sqrt{13};$ that is, $18 = a^3 + 39 a b^2$ and $5 = 3a^2 b + 13 b^3.$ Hard to solve directly, but we can multiply the first by $5$ and subtract $18$ times the second to get

$0 = 5a^3 - 54 a^2 b + 195 a b^2 - 234 b^3,$

which your friendly computer algebra system (otherwise useless) will tell you equals $(a - 3b)(5 a^2 - 39 a b + 78 b^2).$ If $a = 3b,$ then $18 = 144 b^3$ and $5 = 40 b^3,$ so $b = 1/2$ and $a = 3/2$ and

$\displaystyle\sqrt[3]{18 + 5 \sqrt{13}} = \frac{3}{2} + \frac{1}{2}\sqrt{13}.$ Same with "$-$". This method works quite generally.

And would not it? With the prior knowledge that $\displaystyle\sqrt[3]{2 \pm \sqrt{5}} = \frac{1 \pm \sqrt{5}}{2}$ the latter result seems not that surprising, one can think. Bruce continues to show with another example that expectations can be easily shuttered in mathematics.

Again, our previous experience may suggest the existence of rational $a$ and $b$ such that

$\sqrt{2 + \sqrt{3}} = a + b \sqrt{3}.$

If we proceed as before, but now squaring both sides of the equation, and equating rational and irrational components, we get $2 = a^2 + 3b^2$ and $1 = 2 ab,$ so

$a^2 + 3b^2 - 2\cdot 2ab = (a - b)(a - 3b) = 0.$

If $a = b,$ then $2 = 4a^2,$ so $\displaystyle a = \frac{1}{\sqrt{2}} = b.$ This is not what we expected, but is at least true:

$\displaystyle 2 + \sqrt{3} = (\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}})^2 = \frac{1}{2} (1 + \sqrt{3})^2.$

What about the other factor? If $a = 3b,$ then $2 = 12b^2,$ so $\displaystyle b = \frac{1}{\sqrt{6}}$ and $\displaystyle a = \frac{3}{\sqrt{6}}.$ This gives

$\displaystyle 2 + \sqrt{3} = (\frac{3}{\sqrt{6}} + \frac{\sqrt{3}}{\sqrt{6}})^2.$

Strangely, this reduces to exactly the result due to the first (and different) factor:

$\displaystyle \frac{3}{\sqrt{6}} + \frac{\sqrt{3}}{\sqrt{6}} = \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}}.$

In terms of quadratic fields, $\mathbb{Q}[\sqrt{m}]$, what we found may be expressed formally as $\sqrt{2 + \sqrt{3}}\not\in\mathbb{Q}[\sqrt{3}]$ but $\sqrt{2 + \sqrt{3}}\in\mathbb{Q}[\sqrt{2},\sqrt{3}]$ which, perhaps, helps enhance intuition of what goes on in Bruce's example but also adds to the mystery of $\sqrt[3]{18 + 5 \sqrt{13}}\in\mathbb{Q}[\sqrt{13}]$ or $\sqrt[3]{2 + \sqrt{5}}\in\mathbb{Q}[\sqrt{5}].$