Alternate Ending

$\begin{aligned} S & = \frac 1{1^3} - \frac 2{1^3+2^3} + \frac 3{1^3+2^3+3^3} - \frac 4{1^3+2^3+3^3+4^3} + ... \\ & = \sum_{k=1}^\infty \frac {(-1)^{k+1}k}{\sum_{i=1}^k i^3} = \sum_{k=1}^\infty \frac {(-1)^{k+1}k}{\frac {k^2(k+1)^2}{2^2}} = \sum_{k=1}^\infty \frac {(-1)^{k+1}4}{k(k+1)^2} & \small \color{#3D99F6}{\text{Expanding into partial fractions,}} \\ & = 4 \sum_{k=1}^\infty (-1)^{k+1} \left(\frac 1{k} - \frac 1{k+1} - \frac 1{(k+1)^2} \right) \\ & = 8 \color{#3D99F6}{\sum_{k=1}^\infty \frac {(-1)^{k+1}}k} - 4 + 4\color{#D61F06}{\sum_{k=1}^\infty \frac {(-1)^{k+1}}{k^2}} - 4 \\ & = 8 \color{#3D99F6}{\ln 2} + 4 \cdot \color{#D61F06}{\frac 12 \zeta (2)} - 8 & \small \color{#D61F06}{\text{Riemann zeta function }\zeta(2) = \frac {\pi^2}6} \\ & = 8\ln 2 + \frac {\pi^2}3 - 8 \end{aligned}$

$\implies A+B+C = 2+3+8 = \boxed{13}$