$\int_{0}^{\frac{\pi}{6}}\dfrac{\sqrt[3]{\cos x-\cos^3x}}{\cos^3x}dx$
$=\int_{0}^{\frac{\pi}{6}}\dfrac{\sqrt[3]{\cos x-\cos^3x}}{\cos x}d(\tan x)$
$=\int_{0}^{\frac{\pi}{6}}\dfrac{\sqrt[3]{\cos x-\cos^3x}}{\sqrt[3]{\cos^3x}}d(\tan x)$
$=\int_{0}^{\frac{\pi}{6}}\sqrt[3]{\dfrac{\cos x}{\cos^3x}-\dfrac{\cos^3x}{\cos^3x}}d(\tan x)$
$=\int_{0}^{\frac{\pi}{6}}\sqrt[3]{\dfrac{1}{\cos^2x}-1}d(\tan x)$
$=\int_{0}^{\frac{\pi}{6}}\sqrt[3]{1+\tan^2x-1}d(\tan x)$
$=\int_{0}^{\frac{\pi}{6}}\sqrt[3]{\tan^2x}d(\tan x)$
$=\dfrac{3(\tan x)^{\frac{5}{3}}}{5}\left|\begin{array}{l}\dfrac{\pi}{3}\\0\end{array}\right.=\dfrac{\sqrt[6]{3}}{5}$