If $a = 4$, $b = 2$, and $c = -5$, then what is the value of $\sqrt[3]{4a^4b^5} + \dfrac{a - c}{(b+c)^2}$?
+592 Plug in the values:
\(\sqrt[3]{4a^4b^5} + \frac{a-c}{(b+c)^2}\)
\(= \sqrt[3]{4(4)^4(2)^5} + \frac{(4)-(-5)}{((2)+(-5))^2}\)
\(= \sqrt[3]{1024 \cdot 32} + \frac{9}{(-3)^2}\)
\(= \sqrt[3]{32768} + \frac{9}{9}\)
\(= 32 + 1\)
\(= 33\)
