I would want a match for a typographical error (I think!). In a formula in the work of “CARL FRIEDRICH GAUSS, WERKE BAND III. GÖTTINGEN 1866” (www.archive.org).
In chapter “ZUR THEORIE DER TRANSSCENDENTEN FUNCTIONEN GEHÖRIG. [II.][7.]” at page 442 line 13, You read:
$$\frac {A} {a} =\frac {3t-t^{3}+T(1-3tt)} {3[T(1-tt)+t(1-TT)]}.$$
In R.H.S. denominator, the plus sign must be minus as:
$$\frac {A} {a}=\frac {3t-t^{3}+T(1-3tt)} {3[T(1-tt)-t(1-TT)]}.\tag1$$
Question: Why?
Answer: The (1)-equation is a third order modular equation.
Setting
$$T^{2}=(1+p)^{3}\big(\frac {1-p} {1+2p}\big)$$
$$t^{2}=(1+p)\big(\frac {1-p} {1+2p}\big)^{3}$$
with
$$(1+p)^{4}=\frac {T^{3}} {t}.$$
Replacing in that equation we have
$$\frac {1} {m}=\frac {3} {3(1+2p)}$$
therefore
$$m=1+2p.$$
Of course the multiplier of third order equation is
$$m=\frac {a} {A}$$
and $\alpha$ and $\beta$ moduli of first and third degree are related with
$$p^{8}=\frac {\beta^{3}} {\alpha}.$$
The correct version of the Gauss formula is
$$ \frac{A}a = \frac{3t-t^{3}+T(1-3tt)} {3[T(1-tt)-t(1-TT)]}. \tag1 $$
The formula an be simplified to get
$$ \frac{A}a = \frac{t(3-tt)+T(1-3tt)} {3(T - t) (1 + t T)}. \tag2 $$
There are at least six different formulas such as
$$ \frac{A}a = \frac{(2tT)^2-(t+T)^2} {(tt-3TT)+tT(6-4TT)}. \tag3 $$
The generating function of OEIS sequence A261320 is
$$ \frac{A}a = \frac{\theta_3(0,q^3)^2}{\theta_3(0,q)^2} = 1 - 4q + 12q^2 - 28q^3 + 60q^4 + \cdots. \tag4 $$