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Let ${\Phi }_n\left(q\right)$ denote the $n$th cyclotomic polynomial in $q$. Recently, Guo, Schlosser and Zudilin proved that for any integer $n>1$ with $n\equiv 1(mod4)$, $$\sum _{k=0}^{n-1}\frac{{(q^{-1};q^2)}_k^2{(q^{-2};q^4)}_k}{{(q^2;q^2)}_k^2{(q^4;q^4)}_k}{q}^{6k}\equiv 0(mod{\Phi }_n{\left(q\right)}^2),$$ where ${(a;q)}_m=(1-a)(1-aq)\cdots (1-a{q}^{m-1})$. In this note, we give a generalization of the above $q$-congruence to the modulus ${\Phi }_n{\left(q\right)}^3$ case. Meanwhile, we give a corresponding $q$-congruence modulo ${\Phi }_n{\left(q\right)}^2$ for $n\equiv 3(mod4)$. Our proof is based on the ‘creative microscoping’ method, recently developed by Guo and Zudilin, and a ${}_4{\varphi }_3$ summation formula.