<s>[INST] Show that $\sqrt[3]{3\sqrt{21} + 8} - \sqrt[3]{3\sqrt{21} - 8} = 1$ 
Show that
  $$\sqrt[3]{3\sqrt{21} + 8} - \sqrt[3]{3\sqrt{21} - 8} = 1$$

Playing around with the expression, I found a proof which I will post as an answer.
I'm asking this question because I would like to see if there are alternative solutions which are perhaps faster / more direct / elementary / elegant / methodical / insightful etc.
 [/INST] Sometimes you can get lucky de-nesting a radical. This is one of those times.

Write $\sqrt[3]{3\sqrt{21}\pm 8}$ as $\frac12\;\sqrt[3]{24\sqrt{21}\pm 64}$, and consider expressing the radicand as a perfect cube:
$$24\sqrt{21}\pm 64 = (\;p + q \sqrt{21}\;)^3 =p^3 + 3 p^2 q \sqrt{21} + 63 p q^2 + 21 q^3 \sqrt{21}$$
so that
$$p\left(\; p^2 + 63 q^2 \;\right) = \pm 64 \qquad q \left(\;p^2 + 7 q^2\;\right)\cdot 3\sqrt{21} = 8 \cdot 3\sqrt{21}$$
Clearly, we can take $p = \pm 1$ and $q = 1$. Then,
$$\begin{align}
\sqrt[3]{3\sqrt{21}+8} - \sqrt[3]{3\sqrt{21}-8} 
&= \frac{1}{2}\left(\; \sqrt[3]{(1 + \sqrt{21} )^3} - \sqrt[3]{(-1 + \sqrt{21} )^3} \;\right) \\[6pt]
&= \frac{1}{2}\left(\;1 + \sqrt{21} - (-1 + \sqrt{21})\;\right) \\[6pt]
&= 1
\end{align}$$
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