Prove ${_2F_1}\left({{\tfrac16,\tfrac23}\atop{\tfrac56}}\middle|\,\frac{80}{81}\right)=\frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$

My question is related to this question by Vladimir. Because it is already proved that $${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{\small2/3}} = \frac{3^{\small3/2}}{2^{\small4/3}\,5^{\small5/6}\,\pi }\Gamma^3\!\!\left(\tfrac13\right),$$

we have the answer for my question too, since according to Maple

$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{\small2/3}} = \frac{2}{9} \frac{\sqrt[3]{4}\,\sqrt[3]{3}\,\pi^2\,{_2F_1}\left(\begin{array}c\tfrac16,\tfrac23\\\tfrac56\end{array}\middle|\,\frac{80}{81}\right)}{\Gamma^3 \left( \frac{2}{3} \right)}.$$

The second part of the question $-$ the part under the line $-$ is still open.

(This may not be complete, but generalizes the observation for context.) The OP's answer mentioned Reshetnikov's integral which can be expressed as, $$\frac{1}{48^{1/4}\,K(k_3)}\,\int_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt{5}\,x}\,\left(1-x^2\right)^{\small2/3} } =\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-4\big)= \frac{3}{5^{5/6}} $$ so it will be fruitful to find a transformation between Reshetnikov's use of $\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};z_1\big)$ and the OP's use of $\,_2F_1\big(\tfrac{1}{6},\tfrac{2}{3};\tfrac{5}{6};z_2\big)$. Employing well-known transformations, we get,

$$G(y)=\,_2F_1\big(\tfrac{1}{3}, \tfrac{1}{3};\tfrac{5}{6}; -y\big) = \big(\tfrac1{1+2y}\big)^{1/3}\,_2F_1\Big(\tfrac{1}{6}, \tfrac{2}{3};\tfrac{5}{6}; \tfrac{(1+2y)^2-1}{(1+2y)^2}\Big)$$

It is conjectured that if $y$ are certain algebraic numbers, than $G(y)$ is also an algebraic number. Specializing the RHS, define as in this post,

$$H(\tau)=\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;(1-2\delta_4)^2\big),\quad\text{where}\;\frac1{\delta_4}-1=\frac1{27}\left(\tfrac{\eta\big((\tau+1)/3\big)}{\eta(\tau)}\right)^{12}$$ with Dedekind eta function $\eta(\tau)$. If $\tau = \frac{1+N\sqrt{-3}}2$ for some integer $N>1$, then the conjecture implies both $\delta_4$ and $H(\delta_4)$ are algebraic numbers. For example, if we use $\tau = \frac{1+5\sqrt{-3}}2$ and $\tau = \frac{1+7\sqrt{-3}}2$, respectively, then we recover your, $$H\big(\tfrac{1+\color{blue}5\sqrt{-3}}2\big)=\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;\tfrac{80}{81}\big) = \tfrac3{\color{blue}5}\,(9\sqrt5)^{1/3}$$ as well as, $$H\big(\tfrac{1+\color{blue}7\sqrt{-3}}2\big)=\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;\tfrac{3024}{3025}\big) = \tfrac4{\color{blue}7}\,55^{1/3}$$ and so on.

The Tito Piezas answer is related to this parameterization for ${}_2F_1(\frac{1}{6},\frac{2}{3};\frac{5}{6};z)$ in terms of modular forms: $$ {}_2F_1\left(\frac{1}{6},\frac{2}{3};\frac{5}{6}; \frac{j_{3A}}{j_{3A}-108}\right) = \eta(\tau)\; \eta(3\tau)\;(j_{3A}-108)^{1/6}\; \Gamma\left(\frac{2}{3}\right)^3 \frac{(-6i\tau+\sqrt{3}-3i)}{2^{7/3}\pi} \tag{1}$$ where $\eta$ is the Dedekind eta function and $j_{3A}$ is $$ j_{3A} = \left(\frac{\eta(\tau)^6}{\eta(3\tau)^6}+ \frac{3^3 \eta(3\tau)^6}{\eta(\tau)^6}\right)^2 $$ (Choosing the proper branches of the sixth root and the ${}_2F_1$).

Take $\tau = \frac{1}{2} + i \frac{5}{6}\sqrt{3}$ to get the required result from special values of $\eta$...

$$ \eta\left(\frac{1}{2} + i \frac{5}{6}\sqrt{3}\right) =2^{-3/2} 3^{3/8} 5^{-5/12} (1+\sqrt5)^{1/2} \pi^{-1} \Gamma\left(\frac{1}{3}\right)^{3/2} (-1)^{1/24} $$ and something similar for $\eta(3\tau)$, so $$ j_{3A}(\tau) = -8640,\qquad \frac{j_{3A}(\tau)}{j_{3A}(\tau)-108} = \frac{80}{81} $$ In $(1)$, the product of $\Gamma(1/3)$ and $\Gamma(2/3)$ simplify to something algebraic together with a $\pi$, and the powers of $\pi$ cancel.