<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] Here is my solution:
Let $\alpha = \sqrt[3]{3\sqrt{21} + 8}$ and $\beta = \sqrt[3]{3\sqrt{21} - 8}$.
Then
$$ \alpha\beta = \sqrt[3]{(3\sqrt{21} + 8)(3\sqrt{21} - 8)} = \sqrt[3]{(3\sqrt{21})^2 - 8^2} = \sqrt[3]{189 - 64} = \sqrt[3]{125} = 5
$$
and
$$
\alpha^3 - \beta^3 = (3\sqrt{21} + 8) - (3\sqrt{21} - 8) = 16
$$
Now
$$ (\alpha - \beta)^3 = \alpha^3 - 3\alpha^2\beta + 3\alpha\beta^2 - \beta^3 = (\alpha^3 - \beta^3) - 3\alpha\beta (\alpha - \beta) = 16 - 15(\alpha - \beta)
$$
so $\alpha - \beta$ is a root of the polynomial
$$
x^3 + 15x - 16 = (x-1)(x^2 + x + 16).
$$
The part $x^2 + x + 16$ does not have any real roots since its discriminant is $-1 - 4\cdot 16 < 0$.
So $\alpha - \beta$ is a root of $x-1$ and thus
$$
\alpha - \beta = 1
$$
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