How to prove that $\frac{\eta^{14}(q^4)}{\eta^{4}(q^8)}=4\eta^4(q^2)\eta^2(q^4)\eta^4(q^8)+\eta^4(q)\eta^2(q^2)\eta^4(q^4)$? How can we prove that
$$\frac{\eta^{14}(q^4)}{\eta^{4}(q^8)}=4\eta^4(q^2)\eta^2(q^4)\eta^4(q^8)+\eta^4(q)\eta^2(q^2)\eta^4(q^4) \ ?$$
Here, $\eta(q)$ is the Dedekind Eta Function, which is defined by
$$\eta(q)=q^{1/24}\prod_{n=1}^\infty \left(1-q^{n} \right) $$
I looked up some identities of Dedekind Eta Function in the hope of simplifying the right hand expression but I failed. I would be grateful if any one could suggest a proof of this identity.
Source
I came across this identity in a paper. They refer to Entry $t_{8,12,48}$ and $t_{8,18,60a}$ of Dedekind eta function product identities. Unfortunately, I was unable to find these entries on that site.

I will give another answer here. My idea is to use the dimension formula for a space of modular forms with given weight and level.
First, to find the weight and level of the left-hand side, I will use a result of Gordon, Hughes, and Newman for the modularity of some eta-quotients. I do not write down the statement here. You can find this result in Ken Ono's book 'The Web of Modularity' (cf. Theorem 1.64). According to this result, the eta-quotient in the left-hand side is a modular form of weight 5 for $\Gamma_{0}(8)$ with character $\chi$, where
$$\chi(n):=\begin{cases}
1 &\text{if}~n\equiv 1\mod{4},\\
-1 &\text{if}~n\equiv 3\mod{4}.
\end{cases}$$
Note that there is a simple formula for vanishing order of an eta-quotient at cusps (cf. Theorem 1.65 of 'The Web of Modularity'). Using this formula, we see that the left-hand side is a cusp form. We denote by $S_{5}(\Gamma_{0}(8),\chi)$ the space of cusp forms of weight 5 for $\Gamma_{0}(8)$ with character $\chi$ as usual.
Using the dimension formula given by Cohen and Oesterle, we obtain $\dim_{\mathbb{C}}S_{5}(\Gamma_{0}(8),\chi)=2$. From theorems 1.64 and 1.65 in Ken Ono's book, we have that $F_{1}:=\eta^{4}(q^{2})\eta^{2}(q^{4})\eta^{4}(q^{8})$ and $F_{2}:=\eta^{4}(q)\eta^{2}(q^{2})\eta^{4}(q^{4})$ are contained in $S_{5}(\Gamma_{0}(8),\chi)$. Note that $F_{1}$ and $F_{2}$ have Fourier expansions starting with $q^{2}$ and $q$, respectively. This follows easily from the infinite product expansion of $\eta$. Thus, $F_{1}$ and $F_{2}$ are linearly independent, hence forms a basis for $S_{5}(\Gamma_{0}(8),\chi)$. By comparing Fourier coefficients (we can compute a finite number of Fourier coefficients of an eta-quotients using Sage), we have
$$\frac{\eta^{14}(q^{4})}{\eta^{4}(q^{8})}=4F_{1}+F_{2}.$$
