imc

$3!\cdot 5!\cdot 7!=(3\cdot 2\cdot 1)\cdot (5\cdot 4\cdot 3\cdot 2\cdot 1)\cdot (7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1)=2^8\cdot 3^4\cdot 5^2\cdot 7^1$ Therefore, a perfect cube that divides $3!\cdot 5!\cdot 7!$ must be in the form $2^a\cdot 3^b\cdot 5^c\cdot 7^d$ where $a$, $b$, $c$, and $d$ are nonnegative multiples 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 $3\cdot 2\cdot 1\cdot 1=\boxed{6}$