Conjecture $_2F_1\left(\frac14,\frac34;\,\frac23;\,\frac13\right)=\frac1{\sqrt{\sqrt{\frac4{\sqrt{2-\sqrt[3]4}}+\sqrt[3]{4}+4}-\sqrt{2-\sqrt[3]4}-2}}$ Using a numerical search on my computer I discovered the following inequality:
$$\left|\,{_2F_1}\left(\frac14,\frac34;\,\frac23;\,\frac13\right)-\rho\,\right|<10^{-20000},\tag1$$
where $\rho$ is the positive root of the polynomial equation
$$12\,\rho^8-12\,\rho^4-8\,\rho^2-1=0,\tag2$$
that can be expressed in radicals:
$$\rho=\frac1{\sqrt{\sqrt{\frac4{\sqrt{2-\sqrt[3]{4\vphantom{\large1}}}}+\sqrt[3]{4}+4}-\sqrt{2-\sqrt[3]4}-2}}.\tag3$$
Based on this inequality I conjecture that the actual difference is the exact zero, i.e.
$$\color{#808080}{_2F_1\left(\frac14,\frac34;\,\frac23;\,\frac13\right)=\rho}.\tag4$$
I looked up in DLMF and MathWorld, but did not find a known special value with exactly these parameters. It also appears that CAS like Maple or Mathematica do not know this identity.

Could you please suggest any ideas how to prove the conjecture $(4)$?

Update: I can propose even more general conjecture:
$$\color{#808080}{27\,(x-1)^2\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;x\right)^8+18\,(x-1)\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;x\right)^4-8\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;x\right)^2=1}$$
 A: Let's start with this Pfaff transformation for $\,a=\frac 14,b=\frac34,c=\frac23$ :
$$\tag{1}_2F_1\left(a,b\;;c\;;z\right)=(1-z)^{-a}\;_2F_1\left(a,c-b\;;c\;;\frac z{z-1}\right)$$
The 'Darboux evaluation' $(42)$ of Vidunas' "Transformations of algebraic Gauss hypergeometric functions" is :
$$\tag{2}_2F_1\left(\frac14,-\frac 1{12};\,\frac23;\,\frac {x(x+4)^3}{4(2x-1)^3}\right)=(1-2x)^{-1/4}$$
Solving $\,\displaystyle\frac {x(x+4)^3}{4(2x-1)^3}=\frac z{z-1}\,$ gives :
$$\tag{3}z=\frac{x(x+4)^3}{(x^2-10x-2)^2} $$
that we will use as :
$$\tag{4}z-1=\frac {4(2x-1)^3}{(x^2-10x-2)^2}$$
while $(1)$ and $(2)$ return :
$$_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)=\left[(z-1)(2x-1)\right]^{-1/4}$$
so that :
$$_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)=\left[\frac{4\,(2x-1)^4}{(x^2-10x-2)^2}\right]^{-1/4}$$
and (up to a minus sign) :
$$\tag{5}_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)^2=-\frac{x^2-10x-2}{2\,(2x-1)^2}$$
and indeed the substitution of $(z-1)$ and $_2F_1()^2$ with $(4)$ and $(5)$ in your formula gives :
$$27\,(z-1)^2\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^8+18\,(z-1)\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^4-8\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^2=1$$
Many other formulae of this kind may be deduced using Vidunas' paper.
A: The January issue of the Notices has an article by Frits Beukers.  There is a criterion to decide whether certain hypergeometric functions are algebraic.  Checking these numbers in Theorem 2 of that article, we find that
$$
{}_2F_1\left(\frac{1}{4},\frac{3}{4};\frac{2}{3};z\right)
$$
is an algebraic function of $z$.$^*$ He also mentions H. A. Schwarz's list of 1873; I did not check, but this is probably in it.
$^*$ Of course Raymond found (much more than) that in his solution.
