Integrate the denominator

Substitute $x^{\frac {1}{3}} = u \Rightarrow dx = 3x^{\frac {2}{3}}du = 3u^{2}du.$ The integral becomes -

$I = 3\int_{11}^{12} \frac {u^{2}du}{u^{3} - u} = 3\int_{11}^{12} \frac {u du}{u^{2} - 1}$

Let $u^{2} - 1 = a \Rightarrow du = \frac {da}{2u}$ .

$\therefore I = \frac {3}{2}\int_{120}^{143} \frac {da}{a} = \frac {3}{2}\ln(a) \Big|_{120}^{143} = \boxed {\frac {3}{2}\ln\left(\frac {143}{120}\right)}$

Hence $\frac {cd}{ab} = 2860.$

$\begin{aligned} I & = \int_{1331}^{1728} \frac {dx}{x-\sqrt[3] x} & \small \color{#3D99F6} \text{Let }u^3 = x \implies 3u^2 du = dx \\ & = \int_{11}^{12} \frac {3u^2}{u^3-u} du \\ & = \int_{11}^{12} \frac {3u}{u^2-1} du \\ & = \int_{11}^{12} \frac {3u}{(u-1)(u+1)} du & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \frac 32 \int_{11}^{12} \left(\frac 1{u-1}+\frac 1{u+1}\right) du \\ & = \frac 32 \bigg[\ln(u-1) + \ln(u+1)\bigg]_{11}^{12} \\ & = \frac 32 \left(\ln \frac {11}{10} + \ln \frac {13}{12} \right) \\ & = \frac 32 \ln \frac {143}{120} \end{aligned}$

$\implies \dfrac {cd}{ab} = \dfrac {143 \times 120}{3 \times 2} = \boxed{2860}$