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.
 A: 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
$$
\alpha - \beta = 1
$$
A: $(\frac {1 \pm \sqrt{21}}2)^3 = 8 \pm 3\sqrt{21}$
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 $\sqrt[3]{x + iy} + \sqrt[3]{x - iy}$?
A: 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,
$$\begin{align}
\sqrt[3]{3\sqrt{21}+8} - \sqrt[3]{3\sqrt{21}-8} 
&= \frac{1}{2}\left(\; \sqrt[3]{(1 + \sqrt{21} )^3} - \sqrt[3]{(-1 + \sqrt{21} )^3} \;\right) \\[6pt]
&= \frac{1}{2}\left(\;1 + \sqrt{21} - (-1 + \sqrt{21})\;\right) \\[6pt]
&= 1
\end{align}$$
A: As a bonus, we can also obtain that 
$$T=\sqrt[3]{3\sqrt{21} + 8} + \sqrt[3]{3\sqrt{21} - 8}=\sqrt{21}. $$
Indeed, we can write $$T^3 = 6\sqrt{21}+6\sqrt{21}.$$ In order to get rid of $\sqrt{21}$ we can write $T = \alpha \sqrt{21}$ to get the equation
$$21\alpha^3 - 15\alpha-6.$$
We have an obvious root $\alpha=1$, and then prove that 
$$\frac{21\alpha^3 - 15\alpha-6}{\alpha-1} = 21\alpha^2 +21\alpha+6$$does not have real roots.
A: HINT: If we can recognize here Cardano's formula, finding the equation and its obvious root $x=1$ is easy.
A: let $\sqrt[3]{\sqrt{a}+b}-\sqrt[3]{\sqrt{a}-b}=1$
so that $a=189$ and $b=8$
$(\sqrt[3]{\sqrt{a}+b}-\sqrt[3]{\sqrt{a}-b})^3=1$
$\sqrt{a}+b-3(\sqrt[3]{(\sqrt{a}+b)^2}\sqrt[3]{(\sqrt{a}-b)})+3(\sqrt[3]
{(\sqrt{a}+b)}\sqrt[3]{(\sqrt{a}-b)^2})-\sqrt{a}+b=1$
$2b-3(\sqrt[3]{(\sqrt{a}+b)^2}\sqrt[3]{(\sqrt{a}-b)})+3(\sqrt[3]{(\sqrt{a}+b)}\sqrt[3]{(\sqrt{a}-b)^2})=1$
$ 2b-3(\sqrt[3]{(\sqrt{a}+b)}\sqrt[3]{(\sqrt{a}-b)})[ \sqrt[3]{\sqrt{a}+b} -\sqrt[3]{\sqrt{a}-b}]=1$
$2b-3(\sqrt[3]{\sqrt{a}+b}.\sqrt[3]{\sqrt{a}-b})(1))=1$
$2b-3\sqrt[3]{a-b^2}=1$
$2*8-3\sqrt[3]{189-8^2}=1\rightarrow 16-15=1$
