3 divides coefficents of this $q$-series

Question

Denote $\phi(q):=\prod_{j\geq1}(1-q^j)$ and let $\xi=e^{\frac{2\pi i}3}$ be a cube root of unity.

Define the sequence $u(n)$ by $$\prod_{n\geq1}\prod_{s=1}^2(1-q^n\xi^{ns})(1-q^{2n}\xi^{ns}) =\sum_{n\geq0}u(n)\,q^n.$$

QUESTION. Is this true? $$\sum_{n\geq0} u(3n+2)q^{3n+2}=3q^2\,\phi^2(q^9)\,\phi^2(q^{18}).$$

Addendum. To help readers, if we let $$f_n(q)=(1 + q^{3n - 1} + q^{2(3n - 1)})(1 + q^{3n - 2} + q^{2(3n-2)}) (1 - q^{3n})^2,$$ then we have $$\prod_{n\geq1}\prod_{s=1}^2(1-q^n\xi^{ns})(1-q^{2n}\xi^{ns}) =\prod_{n\geq1}f_n(q) f_n(q^2).$$

Answer 1

Trivially $$ \prod_{n\geq1} f_n(q)=\frac{\left(q^3;q^3\right)^4_{\infty }}{(q;q)_{\infty } \left(q^9;q^9\right){}_{\infty }}. $$ Denoting $A(q)=(q;q)_{\infty }(q^2;q^2)_{\infty }$, one can see that in order to prove OP's claim it is enough to perform trisection of $A(q)$, which can be found in Michael Somos' article A Multisection of q-Series . Using formulas from section 4 of this article, by simple algebra one can find $$ A(q \xi)\xi^2+{A\left({q}{\xi}^2\right)}{\xi}=-A(q)-3qA(q^9). $$ This reduces OP's claim to verification of an eta-product identity: $$ A\left(q^3\right)^4-\left(3 q A\left(q^9\right)+A(q)\right) A\left(q^9\right) A(q)^2-9 q^2 A\left(q^9\right)^3 A(q)=0. $$ For this purpose, one can consult Somos' table of such identities. With high probability it can be found there. Even if it can not be found, identities for eta-products can be verified automatically using modular machine.

Thus the identity is related to Ramanujan's famous cubic continued fraction, in other words very classical, textbook stuff.