Consider RSA variation with public key $(e,N)$ where we take $e=3$ and $N=pq$ where:
$p \equiv 1 \text{(mod 3)}$ so that we don't have $\text{gcd}(e,(p-1)(q-1)) \neq 1$
$p \equiv 7 \text{(mod 9)}$ so that we can compute a cubic root of x by doing $x^{\frac{p+2}{9}} \equiv 1 \text{(mod p)}$
For decryption take $d_p = \frac{p+2}{9}$ and $d_q = \frac{q+2}{9}$. Compute $x_p = y^{d_p} \text{(mod p)}$ and $x_q = y^{d_q} \text{(mod q)}$. Finally recover $x = CRT(x_p,x_q)$ where $CRT$ is the function that computes the isomorphism given by the Chinese Remainder theorem.
We know that we can have three different cubic roots for each number in a field. So how is it that given $x_p$ and $x_q$ I can compute $x$ with Chinese Remainder Theorem?