Compute the number of perfect squares that are factors of (5!+6!+7!)^3
All positive integers up to 5 are factors of 5! and so factors of 6! and 7! as well. Since the expression is cubed, each of those, squared, will be a perfect square dividing the expression. Also, both 2 and 3 are in 5! so 6 is a factor of 5!, 6!, and 7!. $\displaystyle 6^2$ is also a perfect square dividing the expression.