Breaking RSA using Chinese Remainder Theorem
I guess you are refering to the case where three people have the following public keys : $(n_1, 3)$, $(n_2, 3)$ and $(n_3, 3)$ (yes, $e = 3$, and this attack is one of the reasons we don't use $e = 3$ anymore).

In this case, if the same message $m$ is sent to those three persons, we have the following system :
$$
\left\{
    \begin{array}{ll}
        c_1 = &amp; m^3\mod n_1 \\
        c_2 = &amp; m^3\mod n_2 \\
        c_3 = &amp; m^3\mod n_3 \\
    \end{array}
\right.
$$

Applying the CRT to
$$
\left\{
    \begin{array}{ll}
        c_1 = &amp; x\mod n_1 \\
        c_2 = &amp; x\mod n_2 \\
        c_3 = &amp; x\mod n_3 \\
    \end{array}
\right.
$$
with $x = m^3$ will give you $x = m^3\mod n_1\times n_2\times n_3$

However, we know that $m \lt n_1,\ n_2,\ n_3$ so we have $ m^3 \lt n_1\times n_2\times n_3$ so a simple cubic root will give us the original message. 

If the message is greater than any $n_i$, you wouldn't be able to recover it with a simple cubic root because you would have a modular result.

Today, we usually use $e=2^{16} + 1 = 65537$ which make this attack much more difficult to apply.