The smallest positive integer $x$ for which $1260x=N^3$, where $N$ is an integer, is:

$\textbf{(A)}\ 1050 \qquad \textbf{(B)}\ 1260 \qquad \textbf{(C)}\ 1260^2 \qquad \textbf{(D)}\ 7350 \qquad \textbf{(6)}\ 44100$
Factoring $1260$ results in $2^2 \cdot 3^2 \cdot 5 \cdot 7$.  If an integer $N$ is a perfect cube, then the exponents of all the primes in its [prime factorization](https://artofproblemsolving.com/wiki/index.php/Prime_factorization) are [multiples](https://artofproblemsolving.com/wiki/index.php/Multiples) of 3.  Thus, the smallest positive integer that can be multiplied by $1260$ to result in a perfect cube is $2 \cdot 3 \cdot 5^2 \cdot 7^2 = 7350$, which is answer choice $\boxed{\textbf{(D)}}$.