An identity involving the Pochhammer symbol I need help proving the following identity:
$$\frac{(6n)!}{(3n)!} = 1728^n \left(\frac{1}{6}\right)_n \left(\frac{1}{2}\right)_n \left(\frac{5}{6}\right)_n.$$
Here,
$$(a)_n = a(a + 1)(a + 2) \cdots (a + n - 1), \quad n > 1, \quad (a)_0 = 1,$$
is the Pochhammer symbol. I do not really know how one converts expressions involving factorials to products of the Pochhammer symbols. Is there a general procedure? Any help would be appreciated.

Pochhammer symbols (sometimes) indicate rising factorials, i.e., $n!=(1)_n$ . This is obviously the case here, since the left hand side is never negative, assuming natural n.
$$\bigg(\frac16\bigg)_n=\prod_{k=0}^{n-1}\bigg(\frac16+k\bigg)=\prod_{k=0}^{n-1}\bigg(\frac{6k+1}6\bigg)=6^{-n}\cdot\prod_{k=0}^{n-1}(6k+1)$$
$$\bigg(\frac12\bigg)_n=\prod_{k=0}^{n-1}\bigg(\frac12+k\bigg)=\prod_{k=0}^{n-1}\bigg(\frac{6k+3}6\bigg)=6^{-n}\cdot\prod_{k=0}^{n-1}(6k+3)$$
$$\bigg(\frac56\bigg)_n=\prod_{k=0}^{n-1}\bigg(\frac56+k\bigg)=\prod_{k=0}^{n-1}\bigg(\frac{6k+5}6\bigg)=6^{-n}\cdot\prod_{k=0}^{n-1}(6k+5)$$
Since $1728=12^3$, our product becomes $$2^{3n}\cdot\prod_{k=0}^{n-1}(6k+1)(6k+3)(6k+5)=\dfrac{2^{3n}\cdot(6n)!}{\displaystyle\prod_{k=0}^{n-1}(6k+2)(6k+4)(6k+6)}=$$
$$=\dfrac{2^{3n}\cdot(6n)!}{2^{3n}\cdot\displaystyle\prod_{k=0}^{n-1}(3k+1)(3k+2)(3k+3)}=\dfrac{(6n)!}{(3n)!}$$
