Correctness of RSA variation using CRT to decrypt?
What you have won't necessarily give you the original value of $x$; it'll give you an $x$ where $x^3 \equiv x_{orig}^3$, however, it isn't necessarily the original value.

To demonstrate this, let us take as an example $p=43$ and $q=61$.

Consider $x_{orig}=5$.  Then, $y=125$, and $x_p = y^5 \bmod 43 = 8$ and $x_q = y^7 \bmod 61 = 52$; this gives us a $CRT(8, 52) = 1943$, which is not the original $x$.  Now, $1943^3 \bmod (43*61) = 125$, and so it also does give the ciphertext, it's just not the same plaintext that you started out with.