Special value of hypergeometric function $\, _2F_1\left(a,a+\frac{1}{3};\frac{4}{3}-a;-\frac{1}{8}\right)$ How can one prove $$\, _2F_1\left(a,a+\frac{1}{3};\frac{4}{3}-a;-\frac{1}{8}\right)=\frac{\left(\frac{2}{3}\right)^{3 a} \Gamma \left(\frac{2}{3}-a\right) \Gamma \left(\frac{4}{3}-a\right)}{\Gamma \left(\frac{2}{3}\right) \Gamma \left(\frac{4}{3}-2 a\right)}$$
This originate from an integral of algebraic functions. What exact transformation can be used to prove this identity? I'd like you to give some suggestions. Thank you!

Update:
The following Mathematica commands verify the quartic transformation given in @pisco's answer:
DifferentialRootReduce[Hypergeometric2F1[4 b/3, (4 b + 1)/3, (4 b + 5)/6, x], x]
DifferentialRootReduce[(1 + 8 x)^(-b) Hypergeometric2F1[b/3, (b + 1)/3, (4 b + 5)/6, 64 x (1 - x)^3/(1 + 8 x)^3], x]
Series[Hypergeometric2F1[4 b/3, (4 b + 1)/3, (4 b + 5)/6, x], {x, 0, 2}]
Series[(1 + 8 x)^(-b) Hypergeometric2F1[b/3, (b + 1)/3, (4 b + 5)/6, 64 x (1 - x)^3/(1 + 8 x)^3], {x, 0, 2}]
Limit[(1 + 8 x)^(-b) Hypergeometric2F1[b/3, (b + 1)/3, (4 b + 5)/6, 64 x (1 - x)^3/(1 + 8 x)^3], x -> -1/8, Direction -> -1]

 A: From the transformation ${_2F_1}(a,b,c,z) = (1-z)^{c-a-b}{_2F_1}(c-a,c-b,c,z)$, it is equivalent to find
$${_2F_1}(\frac{4}{3}-2a,1-2a,\frac{4}{3}-a,-\frac{1}{8}) = {_2F_1}(\frac{4b}{3},\frac{4b+1}{3},\frac{4b+5}{6},-\frac{1}{8})\qquad b=\frac{3}{4}-\frac{3a}{2}$$
We have the quartic transformation:
$$\tag{1}{_2F_1}(\frac{{4b}}{3},\frac{{4b + 1}}{3},\frac{{4b + 5}}{6},x) = {(1 + 8x)^{ - b}}{_2F_1}(\frac{b}{3},\frac{{b + 1}}{3},\frac{{4b + 5}}{6},\frac{{64x{{(1 - x)}^3}}}{{{{(1 + 8x)}^3}}})$$
It holds in a neighborhood of $x=0$ (more precisely $-1/8<x<\frac{3 \sqrt{3}-5}{4} \approx 0.049$). As $x\to -1/8^-$, the argument of $_2F_1$ on RHS tends to $-\infty$. Invoking the asymptotic expansion of ${_2F_1}$ gives
$${_2F_1}(\frac{{4b}}{3},\frac{{4b + 1}}{3},\frac{{4b + 5}}{6},-\frac{1}{8})= (\frac{4}{9})^b \frac{\Gamma (\frac{1}{3}) \Gamma (\frac{4b+5}{6})}{\Gamma (\frac{b}{3}+\frac{5}{6}) \Gamma (\frac{b+1}{3})}$$
so OP's original formula is proved.

Similar to what I said here, $(1)$ is trivial to prove after it has been explicitly conjectured.
