Normally, the chinese theorem looks like this:$$ x = a_1 \bmod n_1\\ x = a_2 \bmod n_2\\ x = a_3 \bmod n_3$$ but what to do if in my situation it looks like this:$$ c_1 = x^e \bmod n_1\\ c_2 = x^e \bmod n_2\\ c_3 = x^e \bmod n_3$$
I know $c_1$, $c_2$, $c_3$, $e$, $n_1$, $n_2$ and $n_3$. So how do I find $x$?
It is the same format. You can arrange the equations as $x^e \equiv c_i $ $mod$ $n_i $ by swapping sides, then you can apply Chinese Remainder Theorem to find the value of $x^e \equiv C $ $mod$ $n_1$$n_2$$n_3$. After that, you may try the cube root attack if public exponent $(e)$ is a small value.
NB_1907 gave the answer for small $e$ (3 if you are given three equations).
Now, let us look at the case where $e$ is potentially large.
In that case, what you end up needing to do is convert it into a series of conventional equivalences, that is:
From $c_1 = x^e \bmod n_1$, you find the value of $x \bmod n_1$ that satisfices it; if $e$ and $n_1$ are relatively prime, the sole value is $n_1 = c_1^{e^-1 \bmod \phi(n_1)}$ [1].
From $c_2 = x^e \bmod n_2$, you similarly find the value of $x \bmod n_2$
From $c_3 = x^e \bmod n_3$, you similarly find the value of $x \bmod n_3$.
From $x \bmod n_1, x \bmod n_2, x \bmod n_3$, you use the CRT conventionally to recover $x \bmod \text{lcm}( n_1, n_2, n_3 )$ (assuming, of course, if it exists)
Now, this is not a great help if you're giving three RSA encryptions of the same value (using the same exponent); it would be easier just to break one of the RSA public keys (rather than all three). However it might be of use if you're given a smaller problem, where the $x$ value might be larger than any of the moduli.
[1]: To determine $\phi(n_1)$, you'll need to factor $n_1$ as a substep.
