How can I prove this closed form for $\sum_{n=1}^\infty\frac{(4n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}$ How can I prove the following conjectured identity?
$$\mathcal{S}=\sum_{n=1}^\infty\frac{(4\,n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}\stackrel?=\frac{\sqrt3}{2\,\pi}\left(2\sqrt{\frac8{\sqrt\alpha}-\alpha}-2\sqrt\alpha-3\right),$$
where
$$\alpha=2\sqrt[3]{1+\sqrt2}-\frac2{\sqrt[3]{1+\sqrt2}}.$$
The conjecture is equivalent to saying that $\pi\,\mathcal{S}$ is the root of the polynomial
$$256 x^8-6912 x^6-814752 x^4-13364784 x^2+531441,$$
belonging to the interval $-1<x<0$.

The summand came as a solution to the recurrence relation
$$\begin{cases}a(1)=-\frac{81\sqrt3}{512\,\pi}\\\\a(n+1)=-\frac{9\,(2n+1)(4n+1)(4 n+3)}{32\,(n+1)(3n+2)(3n+4)}a(n)\end{cases}.$$
The conjectured closed form was found using computer based on results of numerical summation. The approximate numeric result is $\mathcal{S}=-0.06339748327393640606333225108136874...$ (click to see 1000 digits).

This is a very interesting question. Since it is different from similar question of this kind in  that a new technique is involved I will provide the answer. We have:
\begin{eqnarray}
&&\sum\limits_{n=1}^\infty \frac{(4n)!}{\Gamma(n+\frac{2}{3}) \Gamma(n+\frac{4}{3})(n!)^2 (-256)^n}=\\
&&
\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!} \sum\limits_{n=1}^\infty \binom{4 n}{n} \cdot \underbrace{\frac{1}{n+\frac{1}{3}}}_{\int\limits_0^1\theta^{n-\frac{2}{3}} d\theta}\cdot \left( -\frac{3^3}{4^4}\right)^n=\\
&&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\int\limits_0^1 \theta^{-\frac{2}{3}} \cdot \left\{ \frac{x[\theta](1+\theta \frac{27}{256}x[\theta]^3)}{1+4\theta \frac{27}{256}x[\theta]^3}-1\right\} d\theta=\\
&&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\left\{-\frac{4 \cdot 2^{\frac{2}{3}}}{3}\cdot \int\limits_1^{x[1]} \frac{1}{x^{\frac{4}{3}}} \cdot \frac{1}{(1-x)^{\frac{2}{3}}}dx-3\right\}\\
&&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\left\{4 \cdot 2^{\frac{2}{3}}(\frac{1-x[1]}{x[1]})^{\frac{1}{3}}-3\right\}\\
&&\frac{\sqrt{3}}{2 \pi}\left\{4 \cdot 2^{\frac{2}{3}}(\frac{1-x[1]}{x[1]})^{\frac{1}{3}}-3\right\}\\
\end{eqnarray}
In the first line we used elementary properties of factorials. In the second line we used my answer to Closed form solutions for a family of hypergeometric sums. .
Here the function$x[\theta]$ is defined as a solution to the following polynomial equation:
\begin{equation}
1-x[\theta] - \theta \frac{27}{256} x[\theta]^4=0
\end{equation}
where out of the four different solutions we take the solution that is the closest to unity. In the third line we substituted for $x=x[\theta]$, in the fourth line we evaluated the integral and finally in the fifthe line we simplified the result.
Here :
\begin{equation}
x[1]=\frac{2 \sqrt{2+\left(1+\sqrt{2}\right)^{2/3} \left(4 \sqrt{\frac{2}{1+\sqrt{2}-\sqrt[3]{1+\sqrt{2}}}}-2\right)}-2 \sqrt{2 \left(\left(1+\sqrt{2}\right)^{2/3}-1\right)}}{3 \sqrt[6]{1+\sqrt{2}}}
\end{equation}
