The value of 1^3 - 2^3 + 3^3 - + 15^3 is:

The given series is $S = 1^3 - 2^3 + 3^3 - 4^3 + \ldots + 15^3$.

This can be rewritten as the sum of all cubes up to $15$ minus twice the sum of cubes of even numbers up to $14$:

$S = (1^3 + 2^3 + 3^3 + \ldots + 15^3) - 2(2^3 + 4^3 + 6^3 + \ldots + 14^3)$

$S = \sum_{k=1}^{15} k^3 - 2 \times 2^3 (1^3 + 2^3 + 3^3 + \ldots + 7^3)$

$S = \sum_{k=1}^{15} k^3 - 16 \sum_{k=1}^{7} k^3$

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

$\sum_{k=1}^{15} k^3 = \left( \dfrac{15 \times 16}{2} \right)^2 = (120)^2 = 14400$

$\sum_{k=1}^{7} k^3 = \left( \dfrac{7 \times 8}{2} \right)^2 = (28)^2 = 784$

Substituting these values back into the expression for $S$:

$S = 14400 - 16 \times 784$

$S = 14400 - 12544 = 1856$

Answer: $1856$