# Prove this identity using Jacobi's Triple Product Identity

Using Jacobi's Triple Product Identity prove that\begin{align*} \prod_{n \geq 1}\left(1-x^{n}\right)^{6}=&\ \frac{1}{2}\left\{\prod_{n \geq 1}\left(1+x^{2 n-1}\right)^{2}\left(1-x^{2 n}\right)\times\left(1+4 x \frac{d}{d x}\right) 2 \prod_{n>1}\left(1+x^{2 n}\right)^{2}\left(1-x^{2 n}\right)\right.\\ & \left.-2 \prod_{n>1}\left(1+x^{2 n}\right)^{2}\left(1-x^{2 n}\right) \times 4 x \frac{d}{d x} \prod_{n \geqslant 1}\left(1+x^{2 n-1}\right)^{2}\left(1-x^{2 n}\right)\right\} \end{align*}

I found this identity in the A Simple proof of Jacobi's Four Square Theorem. I have given a picture. See that after the step

\begin{align*}\prod_{n\geq 1}(1-x^n)^6\ & =\frac{1}{2}\Bigg\{ \sum_{s=-\infty}^{\infty}x^{s^2} \sum_{r=-\infty}^{\infty} (2r+1)^2x^{r^2+r}-\sum_{r=-\infty}^{\infty}x^{r^2+r}\sum_{r=-\infty}^{\infty}(2s)^2x^{s^2} \Bigg\}\\ & = \frac{1}{2}\Bigg\{ \sum_{s=-\infty}^{\infty}x^{s^2} \times\bigg ( 1+4x\frac{d}{dx}\bigg)\sum_{r=-\infty}^{\infty} x^{r^2+r}-\sum_{r=-\infty}^{\infty}x^{r^2+r}\times 4x\frac{d}{dx}\sum_{r=-\infty}^{\infty}x^{s^2} \Bigg\}\end{align*}

How did they got the identity using Jacobi's Triple Product Identity

The Jacobi Triple Product identity states that $$\sum_{n\in\mathbb{Z}} z^nx^{n^2}=\prod_{n=1}^{\infty} (1-x^{2n})(1+zx^{2n-1})(1+z^{-1}x^{2n-1})\tag{1}$$ for all complex numbers $$x, z$$ such that $$z\neq 0,|x|<1$$.
Putting $$z=1$$ we get $$\sum_{n\in\mathbb {Z}} x^{n^2}=\prod_{n=1}^{\infty} (1+x^{2n-1})^2(1-x^{2n})\tag{2}$$ Putting $$z=x$$ in $$(1)$$ we get $$\sum_{n\in\mathbb {Z}} x^{n^2+n}=\prod_{n=1}^{\infty} (1+x^{2n})(1+x^{2n-2})(1-x^{2n})$$ Note that the factor $$(1+x^{2n-2})$$ equals $$2$$ if $$n=1$$ and thus we can write above equation as $$\sum_{n\in\mathbb{Z}} x^{n^2+n}=2\prod_{n=1}^{\infty} (1+x^{2n})^2(1-x^{2n})\tag{3}$$ The paper of M D Hirschhorn uses the identities $$(2),(3)$$ in their derivation. Let me know if there is anything more which is troubling you in that proof.