Question

(27p^3q^6r^-1)^(1/3)

Answer 1

\[(27p{}^{\wedge}3q{}^{\wedge}6r{}^{\wedge}-1){}^{\wedge}(1/3)\text{ = }\sqrt[3]{\left(\frac{27 p^3 q^6}{r}\right)}\text{ = }\frac{3 p q^2}{\sqrt[3]{r}} \]