Find the value of the fraction 
What common fraction is equivalent to $$\frac{\dfrac{1}{1^3}+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\dfrac{1}{7^3}+\cdots}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots} \text{ ?}$$

I didn't see how to relate the numerator to the denominator. We can't find the value of the numerator and denominator easily it seems, so what else should we try?
 A: $$\sum_{n=1}^{\infty }\frac{1}{n^3}=\sum_{n=1}^{\infty }\frac{1}{(2n)^3}+\sum_{n=1}^{\infty }\frac{1}{(2n-1)^3}=\zeta(3)$$
so
$$\sum_{n=1}^{\infty }\frac{1}{(2n-1)^3}=\zeta(3)-\frac{1}{8}\zeta(3)=\frac{7}{8}\zeta(3)$$
$$\dfrac{\dfrac{1}{1^3}+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\dfrac{1}{7^3}+\cdots}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots}=\frac{\frac{7}{8}\zeta(3)}{\zeta(3)}=???$$
A: \begin{align}
& \frac{\dfrac{1}{1^3}+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\dfrac{1}{7^3}+\cdots}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots} \\[12pt] = {} & \frac{ \left( \dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots\right) - \left( \dfrac 1 {2^3} + \dfrac 1 {4^3} + \dfrac 1 {6^3} + \cdots \right)}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+ \cdots} \\[12pt]
 = {} & \frac{ \left( \dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots\right) - \dfrac 1 {2^3} \cdot \left( \dfrac 1 {1^3} + \dfrac 1 {2^3} + \dfrac 1 {3^3} + \cdots \right)}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+ \cdots} \\[12pt]
= {} & \frac{ \left( 1 - \dfrac 1 {2^3} \right) \left( \dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots\right)}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+ \cdots} \\[12pt]
= {} & 1 - \dfrac 1 {2^3}.
\end{align}
