Surprising difference of cube roots

 Show that

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

is an integer.

Solution below.

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

Then we have that $ab = 2$ and $a^3 - b^3 = 20$.

By binomial theorem we also have $$(a-b)^3 = a^3 - b^3 - 3ab(a-b)$$

Thus the given expression ($a - b$) is a root of

$$t^3 + 6t - 20$$

$t = 2$ is the only real root.