Sum

We have

${(c^2-u^2)}^5 = \displaystyle \sum_{n=0}^5 \dbinom 5n {(c^2)}^{5-n} {(-u^2)}^n$

We only need the sum of the numerical coefficients, so we set $c=u=1$ to get

Sum of numerical coefficients $= {(1^2-1^2)}^5 = \boxed{0}$ .

Substitute $1$ to each of the variable then simplify.

$(c^2 - u^2)^5 = (1^2 - 1^2)^5 = 1 - 1 = 0^5 = 0$