Cubic Cycle is Np-Complete HAMCYCLE = {&lt;&quot;G&quot;&gt; : G is a simple undirected
graph that has a Hamiltonian cycle}

CUBIC CYCLE = {&lt;&quot;G&quot;&gt; :G is a simple undirected graph that contains
a simple cycle of length at least |V (G)|^1/3}
The reduction i figured out is :
G has a ham-cycle &lt;==&gt; G has a cubic cycle of a given length(V^1/3).
Prooving ==&gt; part is simple as we have to show that if a cycle of V exists, then in G a cycle of V^1/3 exists.
But i got stuck in proving the other direction.
Any insights would be helpful.
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.