Given that $\sum_{n=1}^{N} n^{3} = \left[\frac{N}{2}(N+1)\right]^{2},$ Find $1^{2}-2^{3}+3^{3}-4^{3}+\dots-(2N)^{3},$ In terms of N, simplifying your answer.

1. Rewrite the alternating series as the difference of two sum expressions.

Let $S = 1^{2}-2^{3}+3^{3}-4^{3}+\dots-(2N)^{3}$.
Since $1^{2} = 1^{3}$, we can rewrite the series to follow a consistent pattern of cubes:
$S = 1^{3}-2^{3}+3^{3}-4^{3}+\dots-(2N)^{3}$

We can express this alternating sum as the sum of all terms up to $2N$, minus twice the sum of the even terms:
$S = \left(\sum_{n=1}^{2N} n^{3}\right) - 2\left(2^{3} + 4^{3} + \dots + (2N)^{3}\right)$
$S = \sum_{n=1}^{2N} n^{3} - 2\sum_{k=1}^{N} (2k)^{3}$

2. Apply the given sum of cubes formula to evaluate the grouped components.

Using the given formula $\sum_{n=1}^{N} n^{3} = \left[\frac{N}{2}(N+1)\right]^{2}$, we evaluate both parts.

For the sum of all terms up to $2N$:
$\sum_{n=1}^{2N} n^{3} = \left[\frac{2N}{2}(2N+1)\right]^{2} = N^{2}(2N+1)^{2}$

For the even terms, we factor out $2^3 = 8$:
$2\sum_{k=1}^{N} (2k)^{3} = 2 \sum_{k=1}^{N} 8k^{3} = 16 \sum_{k=1}^{N} k^{3}$
$= 16 \left[\frac{N}{2}(N+1)\right]^{2} = 16 \left( \frac{N^{2}(N+1)^{2}}{4} \right) = 4N^{2}(N+1)^{2}$

Substituting these back into our expression for $S$:
$S = N^{2}(2N+1)^{2} - 4N^{2}(N+1)^{2}$

3. Simplify the final expression algebraically.

Extract the common factor $N^{2}$ and expand the terms within the brackets:
$S = N^{2}\left[ (2N+1)^{2} - 4(N+1)^{2} \right]$
$S = N^{2}\left[ (4N^{2}+4N+1) - 4(N^{2}+2N+1) \right]$
$S = N^{2}(4N^{2}+4N+1-4N^{2}-8N-4)$
$S = N^{2}(-4N-3)$
$S = -N^{2}(4N+3)$