During my research on extension of the Jacobi Triple Product. , I finally got an interesting conjecture that is similar and extension of of the Jacobi Triple Product for higher terms . I asked a related question 3 years ago. (The link of my question is) . It should be true with strong sense. I need help to disprove or prove my conjecture (1) below. I do not know if it is known theorem or not.
$$[\phi(qh)-\phi(qh^{-1})]\prod\limits_{n=1}^{ \infty }(1-q^{n}h^{n^2})(1-q^{n}h^{-n^2})=\sum\limits_{n = - \infty }^ \infty (-1)^n q^{\frac{n(n+1)}{2}} h^{\frac{n(n+1)(2n+1)}{6}} \tag 1 $$
Where $\phi(q)$ is Euler function and $h=e^{i.x}$ , $x$ is Real Number
$$\phi(q)=\prod\limits_{n=1}^{ \infty }(1-q^{n})=\sum\limits_{n = - \infty }^ \infty (-1)^n q^{\frac{n(3n-1)}{2}} \tag 2$$
if the series are extended in positive powers of $q$ with coefficients that are Laurent polynomials in $h$, I confirmed the Equation (1) for first few terms (I have confirmed $q^3$ terms manually for now) . I do not have math software extend such q series with $h$ variables in it .
Could you please help me to confirm more terms if my conjecture is true for more terms? I will be very appreciated for confirmation for higher terms or disprove for higher terms that the conjecture is false for higher terms.
$$\phi(qh)-\phi(qh^{-1})= q(h^{-1}-h)+q^2(h^{-2}-h^{2})-q^5(h^{-5}-h^{5})+.... \tag 3 $$
$$\prod\limits_{n=1}^{ \infty }(1-q^{n}h^{n^2})(1-q^{n}h^{-n^2})=1-q(h^{-1}+h)+q^2(1-h^{-4}-h^4)+q^3(h^3+h^{-3}+h^5+h^{-5}-h^9-h^{-9})+....\tag 4$$
$$[\phi(qh)-\phi(qh^{-1})]\prod\limits_{n=1}^{ \infty }(1-q^{n}h^{n^2})(1-q^{n}h^{-n^2})=q(h^{-1}-h)-q^3(h^{-5}-h^5)+.... \tag 5$$
$$\sum\limits_{n = - \infty }^ \infty (-1)^n q^{\frac{n(n+1)}{2}} h^{\frac{n(n+1)(2n+1)}{6}}=q(h^{-1}-h)-q^3(h^{-5}-h^5)+.....\tag 6$$
Thanks a lot for helps
