# A Delightful Piece of Mathematics

Prove: $\sqrt{\sqrt{108} +10} - \sqrt{\sqrt{108} - 10} = 2.$

We will proceed as follows:

For all $a, b$, $a^3-b^3=(a-b)(a^2+ab+b^2)=(a-b)(a^2-2ab+b^2+3ab)=(a-b)((a-b)^2+3ab)$.

That is, $a^3-b^3 = (a-b)((a-b)^2+3ab).\quad\quad\quad(1)$

When $a = \sqrt{\sqrt{108}+10}, \quad b=\sqrt{\sqrt{108}-10},$

we have $a^3 = (\sqrt{\sqrt{108}+10})^3 = \sqrt{108}+10, \quad b^3=(\sqrt{\sqrt{108}-10})^3=\sqrt{108}-10.$

It means $a^3-b^3= 20.$

We also have $ab = \sqrt{\sqrt{108}+10}\cdot\sqrt{\sqrt{108}-10}=\sqrt{(\sqrt{108})^2-10^2}=\sqrt{8}=2.$

As a result, (1) becomes $20 = (a-b)((a-b)^2+3\cdot 2).$

Rewrite it as $20 = x (x^2+6)$

where $x$ denotes $a-b$, we see that $a-b=\sqrt{\sqrt{108}+10}-\sqrt{\sqrt{108}-10}$ is a positive root of $20=x(x^2+6).\quad\quad\quad(\star)$

Clearly, $2$ is also a positive root of $20=x(x^2+6)\quad\quad\quad(\star\star)$

since $20=2\cdot(2^2+6).$

Moreover (see Exercise 1), $20=x(x^2+6)$ has only one positive root $\quad\quad\quad(\star\star\star)$

Therefore, it must be true that $\sqrt{\sqrt{108} +10} - \sqrt{\sqrt{108} - 10} = 2.$

Exercise-1 Prove: $20=x(x^2+6)$ has only one positive root.

Exercise-2 Prove: $\sqrt{8+3\sqrt{21}} + \sqrt{8-3\sqrt{21}} = 1$