A Delightful Piece of Mathematics

June 22, 2020July 13, 2020 / Michael Xue

Prove: $\sqrt[3]{\sqrt{108} +10} - \sqrt[3]{\sqrt{108} - 10} = 2.$

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

$a = \sqrt[3]{\sqrt{108}+10}, \quad b=\sqrt[3]{\sqrt{108}-10},$

we have

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

It means

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

We also have

$ab = \sqrt[3]{\sqrt{108}+10}\cdot\sqrt[3]{\sqrt{108}-10}=\sqrt[3]{(\sqrt{108})^2-10^2}=\sqrt[3]{8}=2.$

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

$a-b=\sqrt[3]{\sqrt{108}+10}-\sqrt[3]{\sqrt{108}-10}$ 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

$\sqrt[3]{\sqrt{108} +10} - \sqrt[3]{\sqrt{108} - 10} = 2.$

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

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

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