Find the Value of Alternating Sum of Cubes

To evaluate the alternating sum of cubes: $S = 1^3 - 2^3 + 3^3 - 4^3 + \dots + 15^3$

We can rewrite the series by adding and subtracting the even terms to complete the sum of the first $15$ cubes: $S = (1^3 + 2^3 + 3^3 + \dots + 15^3) - 2(2^3 + 4^3 + 6^3 + \dots + 14^3)$

Using the formula for the sum of the first $n$ cubes, $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$:

1. Calculate the sum of all cubes from $1^3$ to $15^3$: $\sum_{k=1}^{15} k^3 = \left(\frac{15 \times 16}{2}\right)^2 = (120)^2 = 14400$

2. Calculate the sum of the even cubes: $2^3 + 4^3 + 6^3 + \dots + 14^3 = 2^3(1^3 + 2^3 + 3^3 + \dots + 7^3)$ $= 8 \times \left(\frac{7 \times 8}{2}\right)^2 = 8 \times (28)^2 = 8 \times 784 = 6272$

3. Multiply the sum of even cubes by $2$: $2 \times (2^3 + 4^3 + 6^3 + \dots + 14^3) = 2 \times 6272 = 12544$

Now, substitute these back into the expression for $S$: $S = 14400 - 12544 = 1856$

Alternatively, we can use the general formula for the alternating sum of the first $N$ cubes when $N = 2n - 1$ is odd: $S_{2n-1} = \sum_{k=1}^{2n-1} (-1)^{k-1} k^3 = n^2 (4n - 3)$

Here, $2n - 1 = 15 \implies n = 8$.

Substituting $n = 8$: $S_{15} = 8^2 \times (4(8) - 3) = 64 \times (32 - 3) = 64 \times 29 = 1856$

Thus, the correct option is B.