Question

write simplest rationalizing factor of
(i)$\sqrt[3]{36}$

Answer 1

Notice that $\sqrt[3]{36}=\sqrt[3]{6^2}$
and $\sqrt[3]{6^2}\times\sqrt[3]{6}=\sqrt[3]{6^2\times 6}=\sqrt[3]{6^3}=6$

That means $\boxed{~\sqrt[3]{6}~}$ is the simplest rationalising factor of $\sqrt[3]{36}$