Not my calculus problem #3

$\begin{aligned} I & = \int_0^{\pi/4} \frac {\tan^2 x}{\sqrt[3]{\tan x - x}}\ dx & \small \blue{\text{Let }u = \tan x - x} \\ & = \int_0^{1-\pi/4} \frac {du}{u^{1/3}} & \small \blue{\implies du = (\sec^2 x -1) dx = tan^2 x \ dx} \\ & = \frac {3u^{2/3}}2 \ \bigg|_0^{1-\pi/4} \\ & = \frac {3(4-\pi)^{2/3}}{2\cdot 4^{2/3}} \end{aligned}$

Therefore $a+b+c = 3+2+7 = \boxed{12}$ .