Kind of Problem we do in Engg.

$\begin{aligned} I & = \int_0^2 x(8-x^3)^\frac 13 dx \\ & = \int_0^2 2x \sqrt[3]{1-\frac {x^3}8} dx & \small \color{#3D99F6} \text{Let }u = \frac x2 \implies 2 du = dx \\ & = \int_0^1 8u \sqrt[3]{1-u^3} du & \small \color{#3D99F6} \text{Let }t = u^3 \implies dt = 3u^2 du \\ & = \frac 83 \color{#3D99F6} \int_0^1 t^{-\frac 13} (1-t)^\frac 13 dt & \small \color{#3D99F6} \text{Beta function }B(m,n) = \int_0^1 t^{m-1}(1-t)^{n-1} dt \\ & = \frac 83 \color{#3D99F6} B \left(\frac 23, \frac 43\right) & \small \color{#3D99F6} B(m,n) = \frac {\Gamma (m)\Gamma (n)}{\Gamma (m+n)} \\ & = \frac {8\color{#3D99F6} \Gamma \left(\frac 23\right) \Gamma \left(\frac 43\right)}{3\color{#D61F06} \Gamma (2)} & \small \color{#3D99F6} \text{where }\Gamma (\cdot) \text{ denotes the gamma function.} \\ & = \frac 83 \Gamma \left(\frac 23\right) \color{#3D99F6} \Gamma \left(\frac 43\right) & \small \color{#D61F06} \text{Since } \Gamma(n) = (n-1)! \implies \Gamma(2) = 1! = 1 \\ & = \frac 83 \Gamma \left(\frac 23\right) \cdot \color{#3D99F6} \frac 13 \Gamma \left(\frac 13\right) & \small \color{#3D99F6} \text{and }\Gamma(1+s) = s\Gamma (s) \\ & = \frac 89 \color{#3D99F6} \Gamma \left(\frac 13\right) \Gamma \left(\frac 23\right) & \small \color{#3D99F6} \text{also }\Gamma (s) \Gamma(1-s) = \frac \pi{\sin (x\pi)} \\ & = \frac 89 \cdot \frac \pi {\sin \frac \pi 3} = \frac {16 \pi}{9\sqrt 3} \end{aligned}$

Therefore, $a+b+c = 16 + 9 + 3 = \boxed{28}$ .

References:

• Beta function
• Gamma function