jmc

Written as a product of primes, we have $$3!\cdot 5!\cdot 7!=2^8\cdot 3^4\cdot 5^2\cdot 7.$$A cube that is a factor has a prime factorization of the form $2^p\cdot 3^q\cdot 5^r\cdot 7^s$, where $p$, $q$, $r$, and $s$ are all multiples of 3. There are 3 possible values for $p$, which are 0, 3, and 6. There are 2 possible values for $q$, which are $0$ and $3$. The only value for $r$ and for $s$ is 0. Hence, there are $\boxed{6}=3\cdot 2\cdot 1\cdot 1$ distinct cubes that divide $3!\cdot 5!\cdot 7!$. They are

$$\begin{array}{rl}1& =2^0{3}^0{5}^0{7}^0,8=2^3{3}^0{5}^0{7}^0,27=2^0{3}^3{5}^0{7}^0,\\ 64& =2^6{3}^0{5}^0{7}^0,216=2^3{3}^3{5}^0{7}^0,\text{and}1728=2^6{3}^3{5}^0{7}^0.\end{array}$$