A hypergeometric puzzle $$
143\,\sqrt {3}\;{\mbox{$_2$F$_1$}\left(\frac{1}{2},\frac{1}{2};\,1;\,{\frac {3087}{8000}}\right)}=
40\,\sqrt {5}\;
{\mbox{$_2$F$_1$}\left(\frac{1}{3},\frac{2}{3};\,1;\,{\frac {2923235}{2924207}}\right)}
$$
While working on another problem, I found two different ways of expressing my solution.  Plugging in a value, I get the above equation.  
Are there methods/references for proving something like this? 
Maple does not know them, it seems.
added 
More generally, I have
$$
\frac{\displaystyle 3\,\sqrt {2}\;
{\mbox{$_2$F$_1$}\Bigg(\frac{1}{2},\frac{1}{2};\,1;\,{\frac { 64\left( 1-y \right) ^{3}}{ \left( -{y}^{3/2}\sqrt {y+8}-8\,\sqrt {y}\sqrt {y+8}+{y}^{2}-20\,y-8 \right) ^{2}}}}\Bigg)}
{{\sqrt {-3\,{y}^{2}+60\,y+24+3\,{y}^{3/2}\sqrt {y+8}+24\,
\sqrt {y}\sqrt {y+8}}}}
=
\frac{\displaystyle {\mbox{$_2$F$_1$}\bigg(\frac{1}{3},\frac{2}{3};\,1;\,{\frac { \left( y+8 \right) ^{2} \left( 1-y \right) }{ \left( 4-y \right) ^{3}}}\bigg)}}
 {( 4 - y)}
$$
for $0 < y < 4$
... and for the above problem I found a value of $y$ so that both $y$ and $y+8$ were perfect squares.   
But I was hoping someone knew methods to do it other than the round-about way I came to it.
 A: These formulas originate from modular parametrizations of the underlying hypergeometric functions and there are many such appearances in the literature (a collection of references can be found in http://arxiv.org/abs/1302.0548 where such algebraic transformations are used in a related context). But the most classical source is Goursat's treatment of such transformations in his thesis:
http://www.numdam.org/item?id=ASENS_1881_2_10__S3_0.
A: The OP's general identity is only half equivalent to the one on page 258 of Ramanujan's second notebook. (Also Theorem 5.6 of Berndt's Ramanujan's Notebooks Vol 5, p.112.) It needs a second one, namely Theorem 6.1 on p.116 to complete the puzzle pieces.

I. Piece 1: Theorem 5.6 If

$$u_1 = \frac{p^3(2+p)}{1+2p},\quad\quad v_1 = \frac{27p^2(1+p)^2}{4(1+p+p^2)^3}$$
then for $0\leq p <1$,
$$(1+p+p^2)\,_2F_1\big(\tfrac12,\tfrac12;1;u_1\big)=\sqrt{1+2p}\;_2F_1\big(\tfrac13,\tfrac23;1;v_1\big)\tag1$$

II. Piece 2: Theorem 6.1 If

$$u_2 = \frac{q(3+q)^2}{2(1+q)^3},\quad\quad v_2 = \frac{q^2(3+q)}{4}$$
then for $0\leq q <1$,
$$\,_2F_1\big(\tfrac13,\tfrac23;1;u_2\big)=(1+q)\;_2F_1\big(\tfrac13,\tfrac23;1;v_2\big)\tag2$$

III. Piece 3: $v_1=v_2$

It turns out the OP's was the special case $v_1=v_2$, hence,
$$\frac{27p^2(1+p)^2}{4(1+p+p^2)^3} = \frac{q^2(3+q)}{4}$$
While a cubic in $q$, it conveniently factors linearly. Choosing the correct factor such that $0\leq p,q <1$, we can aesthetically express the OP's general identity as,
$$\large \tfrac{(1+p+p^2)(1+q)}{\sqrt{1+2p}}\,_2F_1\Big(\tfrac12,\tfrac12;1;\tfrac{p^3(2+p)}{1+2p}\Big)=\;_2F_1\Big(\tfrac13,\tfrac23;1;\tfrac{q(3+q)^2}{2(1+q)^3}\Big)\tag{3a}$$
where,
$$q = \frac{3p}{1+p+p^2}\tag{3b}$$
For example, the OP used $y=\tfrac1{36}$. But using $p=\tfrac7{10}$ on $(3)$, we recover the original equality,
$$\tfrac{143\sqrt3}{40\sqrt5}\;{\mbox{$_2$F$_1$}\left(\tfrac{1}{2},\tfrac{1}{2};\,1;\,{\tfrac {3087}{8000}}\right)}={\mbox{$_2$F$_1$}\left(\tfrac{1}{3},\tfrac{2}{3};\,1;\,{\tfrac {2923235}{2924207}}\right)}$$
and puzzle solved.
A: This should be related to Ramanujan's theory of cubic elliptic functions. Presumably your identity is equivalent to one occurring on page 258 of Ramanujan's second notebook, see Thm. 5.6 in Berndt, Bhargava and Garvan, Ramanujan's theory of elliptic functions to alternative bases, Trans. Amer. Math. Soc. 347 (1995), 4163-4244. You can download a pdf from Frank Garvan's homepage. 
