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.

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$
$2b-3(\sqrt[3]{\sqrt{a}+b}.\sqrt[3]{\sqrt{a}-b})(1))=1$
$2b-3\sqrt[3]{a-b^2}=1$
$2*8-3\sqrt[3]{189-8^2}=1\rightarrow 16-15=1$
