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.

$(\frac {1 \pm \sqrt{21}}2)^3 = 8 \pm 3\sqrt{21}$

These answers are also relevant I guess Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? How does one evaluate $\sqrt[3]{x + iy} + \sqrt[3]{x - iy}$?

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 $$ \alpha - \beta = 1 $$