A Delightful Piece of Mathematics

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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