Sum Time!

$\begin{aligned} S & = \frac{\left( 1- \frac{1}{2^3}+\frac{1}{3^3}-\frac{1}{4^3}+\frac{1}{5^3}-\ldots\right)}{\left(1+ \frac{1}{2^3}+\frac{1}{3^3}+ \frac{1}{4^3}+\frac{1}{5^3}+\ldots \right)} \\ & =\frac{ \displaystyle \sum^{\infty}_{n=1}\frac{1}{n^3}-\frac{2}{8}\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^3}}{\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^3}} \\ & = \frac{ \displaystyle \sum^{\infty}_{n=1}\frac{1}{n^3}\left(1-\frac{1}{4}\right)}{ \displaystyle \sum^{\infty}_{n=1}\frac{1}{n^3}} \\ & = \frac{3}{4} \\ \Rightarrow 3×4 & = \boxed{12}\end{aligned}$

The Fraction can be re-written as $\frac{\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3}}{\Large \displaystyle \sum_{n=1}^\infty \frac{1}{n^3}}$

Consider this series $-1+\frac{1}{2^3}-\frac{1}{3^3}+\frac{1}{4^3}-\cdots$

To perform the analytic continuation, write $\Bigg(-1+\frac{1}{2^3}-\frac{1}{3^3}+\frac{1}{4^3}-\cdots\Bigg)+\Bigg(1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots\Bigg)$ $=2\Bigg(\frac{1}{2^3}+\frac{1}{4^3}+\frac{1}{6^3}+\cdots\Bigg)$ $=2^{1-3}\Bigg(1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots \Bigg)$ Rewriting in the terms of Riemann Zeta Function, $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}}{n^3}+\zeta(3)=2^{1-3}\zeta(3)$ $\zeta(3)-\frac{1}{4}\zeta(3)=-\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}}{n^3}$ $\Bigg(1-\frac{1}{4}\Bigg)\zeta(3)=\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3}$ Dividing both sides by $\zeta(3)$ gives $\frac{3}{4}=\frac{\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3}}{\zeta(3)}$ $\frac{\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3}}{\Large \displaystyle \sum_{n=1}^\infty \frac{1}{n^3}}=\frac{3}{4}$

So, the answer is $3 \times 4=\boxed{12}$