If $a = 4$, $b = 2$, and $c = -5$, then what is the value of $\sqrt[3]{4a^4b^5} + \dfrac{a - c}{(b+c)^2}$?
$\sqrt[3]{4a^4b^5} + \dfrac{a - c}{(b+c)^2} = 13$
