Cracking plain RSA without private key
It depends on what you mean by "cracking", and also what you mean by "RSA".

"Plain RSA", also sometimes described as "textbook RSA", is the lone modular exponentiation. By itself, it is very weak. For instance, if $n$ is the modulus and $e$ the public exponent, suppose that you want to "decrypt" a cipher text $c_1$ (i.e. you want to find $m_1$ such that $m_1^e = c_1 \pmod n$). You can choose a random integer $c_2$ modulo $n$, and compute $c_3 = c_1/c_2 \pmod n$; then, obtain the decrypted messages $m_2$ and $m_3$, corresponding to $c_2$ and $c_3$, respectively. In that situation, you know that $m_1 = m_2m_3 \pmod n$. This is a chosen ciphertext attack, in which you ask for two decryptions and obtain three plaintext messages, including the target message $m_1$, and yet the decryption oracle did not see any request ending in revealing $m_1$. That's "cracked" by any decent definition of "cracking".

That's why "plain RSA" is not RSA. RSA, the asymmetric encryption algorithm, is described by PKCS#1 and includes a padding operation which is essential to security. With appropriate padding, there is no known attack on RSA which would help in decrypting a given message, even in an adaptative chosen ciphertext context.