Entry 94

Level 4. Given $q = e^{2\pi i \tau}$, Ramanujan's theta function $f(a,b)$ and its one-parameter version as $f(-q) = f(-q,-q^2)$. We have the sum-product identities,

$$\begin{align}A(q) &= \frac{f(-q,-q^3)}{f(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+n}} {(q^2;q^2)_n} = \prod_{n=1}^\infty\frac{\alpha_4 }{(1-q^{4n-2})(1-q^{4n-2})} = q^{-1/12}\frac{\eta(4\tau)}{\eta(2\tau)}\\ B(q) &= \frac{f(-q^2,-q^2)}{f(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}} {(q^2;q^2)_n}  = \prod_{n=1}^\infty\frac{\alpha_4 }{(1-q^{4n-1})(1-q^{4n-3})} = q^{1/24}\frac{\eta^2(2\tau)}{\eta(\tau)\eta(4\tau)}\end{align}$$

where $\alpha_4 = \dfrac{(1-q^{2n})}{(1-q^{4n})}$. 

Let $a = \sqrt2\,q^{1/12}A(q)$ and $b= q^{-1/24}B(q)$, then $(a, b)$ are just eta quotients and for appropriate quadratic $\tau$ are actually radicals. Furthermore, the difference of their $8$th powers is also an $8$th power $$-a^8+b^8 =-\left(\frac{\sqrt2\,\eta(4\tau)\,}{\;\eta(2\tau)}\right)^8+\left(\frac{\eta^2(2\tau)}{\eta(\tau)\eta(4\tau)}\right)^8=\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^8$$