How many positive cubes divide $3! \cdot 5! \cdot 7!$ ?
$\textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 5\qquad \textbf{(E) } 6$
$3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8}\cdot3^{4}\cdot5^{2}\cdot7^{1}$
Therefore, a [perfect cube](https://artofproblemsolving.com/wiki/index.php/Perfect_cube) that divides $3! \cdot 5! \cdot 7!$ must be in the form $2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}$ where $a$, $b$, $c$, and $d$ are [nonnegative](https://artofproblemsolving.com/wiki/index.php/Nonnegative) [multiples](https://artofproblemsolving.com/wiki/index.php/Multiple) of $3$ that are less than or equal to $8, 5, 2$ and $1,$ respectively. 
So: 
$a\in\{0,3,6\}$ ($3$ possibilities)
$b\in\{0,3\}$ ($2$ possibilities)
$c\in\{0\}$ ($1$ possibility)
$d\in\{0\}$($1$ possibility)
So the number of perfect cubes that divide $3! \cdot 5! \cdot 7!$ is $3\cdot2\cdot1\cdot1 = \boxed{\textbf{(E) }6}$