Cubic Cycle is Np-Complete
The reduction that you suggested does not work. For example, consider a graph $G = (V, E)$ that consists of 9 disjoint triangles ( triangles that do not touch each other). $G$ has a triangle which a hamiltonian cycle of length $3 = 27^{\frac{1}{3}} = |V|^{\frac{1}{3}}$, yet $G$ has no hamiltonian cycle.
A simple correct reduction operates as follows. Given an undirected graph $G = (V, E)$, the reduction outputs $G$ with additional $|V|^3 - |V|$ isolated vertices. I leave the details to you.