Complexity of finding 2 vertex-disjoint $(|V|/2)$-cycles in cubic graphs? I posted this on mathoverflow but with no luck:

Finding a connected 2-factor is $NP$-complete since it is equivalent to the Hamiltonian cycle problem. I'm interested in the complexity of finding two vertex-disjoint $(|V|/2)$-cycles in cubic graphs.

I suspect that it is $NP$-complete but I'm not able to find a reference. Also, What is the complexity of finding two vertex-disjoint $(|V|/2)$-cycles with equal cardinality in planar cubic bipartite graphs? Is it $NP$-complete?

Providing references is highly appreciated. $NP$-completeness refers to the decision version of the problem.
I assume, as stated in the comments, that the problem is to find a 2-factor with two cycles of length $|V|/2$ each in a cubic graph.
This problem is NP-complete:

1) HC in cubic graphs is NP-complete.

2) The following problem HC' is also NP-complete: Given a cubic graph $G$ and an edge $e$
in $G$, is there a Hamiltonian cycle through $e$. We can reduce HC to it: Given a graph $G$, replace one arbitrary vertex (left hand side) by a triangle with another vertex in it (right hand side):



Call this new graph $G'$.
$G'$ is Hamiltonian iff $G$ is. Furthermore, if $G'$ is Hamiltonian, then for any red edge, there is a Hamiltonian cycle that contains this red edge. (Without the vertex in the middle, every Hamiltonian cycle contains two green edges and any green edge can then be replaced by two red ones.) The reduction from HC to HC' maps a cubic graph $G$ to $G'$ together with any red ege.

3) Finally, we reduce HC' to the problem of the question. Given an instance
$(G,e)$, we create two copies of $G$, subdivide $e$ in each copy and connected the two vertices that subdivide the two copies by a new edge $f$. This edge $f$ is a bridge in the new graph. If $G$ has a Hamiltonian cycle through $e$, then the new graph has a 2-factor consisting of two cycles of the same size.
Conversely, if the new graph has such a 2-factor, then it cannot use $f$. Therefore, it splits into two Hamiltonian cycles of the two copies. Since in these copies, the vertex subdividing $e$ has degree two, we get a Hamiltonian cycle of $G$ that contains $e$.