The sum $$\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \ldots$$ up to 8 terms, is :

The series has its $$n^{\text{th}}$$ term

$$T_n=\frac{1^3+2^3+\dots+n^3}{1+3+5+\dots+(2n-1)}$$

Step 1 - Evaluate the numerator
The sum of the first $$n$$ cubes is a standard result:
$$1^3+2^3+\dots+n^3=\left[\frac{n(n+1)}{2}\right]^2 \; -(1)$$

Step 2 - Evaluate the denominator
The denominator is the sum of the first $$n$$ odd numbers:
$$1+3+5+\dots+(2n-1)=n^2 \; -(2)$$

Step 3 - Form the $$n^{\text{th}}$$ term
Using $$(1)$$ and $$(2)$$:
$$T_n=\frac{\left[\dfrac{n(n+1)}{2}\right]^2}{n^2}=\frac{(n(n+1))^2}{4n^2}=\frac{(n+1)^2}{4}$$

Step 4 - Sum the first 8 terms
$$S_8=\sum_{n=1}^{8}T_n=\frac14\sum_{n=1}^{8}(n+1)^2$$

Put $$k=n+1\implies k=2,3,\dots,9$$, so
$$\sum_{n=1}^{8}(n+1)^2=\sum_{k=2}^{9}k^2=\left(\sum_{k=1}^{9}k^2\right)-1^2$$

The formula for the sum of squares is $$\sum_{k=1}^{m}k^2=\frac{m(m+1)(2m+1)}{6}$$. For $$m=9$$:
$$\sum_{k=1}^{9}k^2=\frac{9\cdot10\cdot19}{6}=285$$

Therefore,
$$\sum_{k=2}^{9}k^2=285-1=284$$

Hence,
$$S_8=\frac14\times284=71$$

So the required sum is $$71$$.

Option B which is: 71