  Let $q$ be a scalar that is not a root of unity. We show that any polynomial
in the Casimir element of the Fairlie-Odesskii algebra $U_q'(\mathfrak{so}_3)$
cannot be expressed in terms of only Lie algebra operations performed on the
generators $I_1,I_2,I_3$ in the usual presentation of $U_q'(\mathfrak{so}_3)$.
Hence, the vector space sum of the center of $U_q'(\mathfrak{so}_3)$ and the
Lie subalgebra of $U_q'(\mathfrak{so}_3)$ generated by $I_1,I_2,I_3$ is direct.
