Sum of First Eight Terms of Given Cubes Series

To find the sum of the first $8$ terms of the given series, we first determine the $n$-th term, $T_n$.

The given series is: $\frac{1^3}{1} + \frac{1^3+2^3}{1+3} + \frac{1^3+2^3+3^3}{1+3+5} + \dots$

The $n$-th term $T_n$ is given by: $T_n = \frac{1^3 + 2^3 + 3^3 + \dots + n^3}{1 + 3 + 5 + \dots + (2n-1)}$

Using the standard summation formulas:

1. Sum of cubes of the first $n$ natural numbers: $\sum_{k=1}^n k^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4}$

2. Sum of the first $n$ odd natural numbers: $\sum_{k=1}^n (2k-1) = n^2$

Substituting these formulas into $T_n$, we get: $T_n = \frac{\frac{n^2(n+1)^2}{4}}{n^2} = \frac{(n+1)^2}{4}$

Now, we need to calculate the sum of the first $8$ terms, denoted by $S_8$: $S_8 = \sum_{n=1}^8 T_n = \sum_{n=1}^8 \frac{(n+1)^2}{4} = \frac{1}{4} \sum_{n=1}^8 (n+1)^2$

By changing the index of summation with $k = n+1$, the sum becomes: $\sum_{n=1}^8 (n+1)^2 = \sum_{k=2}^9 k^2 = \left( \sum_{k=1}^9 k^2 \right) - 1^2$

Using the formula for the sum of squares of the first $m$ natural numbers, $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$: $\sum_{k=1}^9 k^2 = \frac{9 \times 10 \times 19}{6} = 285$

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

Finally, substituting this value back into $S_8$: $S_8 = \frac{1}{4} \times 284 = 71$

Hence, the correct option is B.