How to derive the closed form of this gamma quotient? Sorry for this very localized question. I am aware that there are myriads of algebraic gamma quotients of this kind around, but in fact I have never seen proofs involving more than the well-known reflection and multiplication formulae of the gamma function. Though something deeper is obviously needed here...  
How to prove this conjectured identity?
$$\boxed{\frac{ \Gamma(\frac{1 }{21})\Gamma(\frac{4 }{21})\Gamma(\frac{16 }{21})}
{ \Gamma(\frac{2 }{21})\Gamma(\frac{8 }{21})\Gamma(\frac{ 11}{21})}\stackrel{\color{red}?}=\sqrt[3\  \ ]{14+3\sqrt{21}}\ }$$
The minimal polynomial (all numerically found) is $x^6-28x^3+7=0$. With that many multiples of $7$ occurring everywhere, I'd guess that there must be a whole family of such identities. Though for the next gamma quotient in the row, which should be
$\frac{ \Gamma(\frac{1 }{85})\Gamma(\frac{4 }{85}) \Gamma(\frac{16 }{85})\Gamma(\frac{64 }{85})}
{ \Gamma(\frac{2 }{85})\Gamma(\frac{8 }{85})\Gamma(\frac{ 32}{85})\Gamma(\frac{ 43}{85})}  $, my computer can't find any algebraic form.
 A: It has been speculated that all such relations of gamma functions can be derived from reflection identity and multiplication theorem. Another example can be found here.
For your special case, omitting the $\Gamma$ sign, we have
$$\begin{aligned}
\frac{{\left( {\frac{1}{{21}}} \right)\left( {\frac{4}{{21}}} \right)\left( {\frac{{16}}{{21}}} \right)}}{{\left( {\frac{2}{{21}}} \right)\left( {\frac{8}{{21}}} \right)\left( {\frac{{11}}{{21}}} \right)}} &= \frac{{\left( {\frac{4}{{21}}} \right)\left( {\frac{{16}}{{21}}} \right)}}{{\left( {\frac{2}{{21}}} \right)\left( {\frac{8}{{21}}} \right)\left( {\frac{{11}}{{21}}} \right)}}{(2\pi )^3}{7^{1/2 - 1/3}}\frac{{\left( {\frac{1}{3}} \right)}}{{\left( {\frac{4}{{21}}} \right)\left( {\frac{1}{3}} \right)\left( {\frac{{10}}{{21}}} \right)\left( {\frac{{13}}{{21}}} \right)\left( {\frac{{16}}{{21}}} \right)\left( {\frac{{19}}{{21}}} \right)}} \\
 &= {(2\pi )^3}{7^{1/6}}\frac{1}{{\left( {\frac{2}{{21}}} \right)\left( {\frac{8}{{21}}} \right)\left( {\frac{{11}}{{21}}} \right)\left( {\frac{{10}}{{21}}} \right)\left( {\frac{{13}}{{21}}} \right)\left( {\frac{{19}}{{21}}} \right)}} \\
& = \frac{{{{(2\pi )}^3}{7^{1/6}}}}{{{\pi ^3}}}\sin \left( {\frac{{2\pi }}{{21}}} \right)\sin \left( {\frac{{8\pi }}{{21}}} \right)\sin \left( {\frac{{10\pi }}{{21}}} \right)\\
&= \sqrt[3\  \ ]{14+3\sqrt{21}} \end{aligned}$$
in the first step, I used $$\Gamma (x)\Gamma (x + \frac{1}{m}) \cdots \Gamma (x + \frac{m-1}{m}) = (2\pi )^{(m-1)/2}{m^{\frac{1}{2} - mx}}\Gamma (mx)$$
in the last step, it is a direct computation on minimal polynomial. 
