2005 AMC 10A Problem 15

As a product of primes, $3! \cdot 5! \cdot 7! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7.$ A cube divisor uses exponents that are multiples of $3:$ the exponent of $2$ can be $0,$ $3,$ or $6$ ($3$ choices), the exponent of $3$ can be $0$ or $3$ ($2$ choices), and the exponents of $5$ and $7$ must be $0.$ That gives $3 \cdot 2 \cdot 1 \cdot 1 = 6$ cubes. Thus, the correct answer is E.