The problem presents the expression:
\[ \left( 1 + x + x^2 + x^3 \right)^5 = \sum_{k=0}^{15} a_k x^k. \]
The goal is to compute:
\[ S = \sum_{k=0}^7 (-1)^k \cdot a_{2k}. \]
Step 1: Simplify the base. Let:
\[ P(x) = 1 + x + x^2 + x^3. \]
Using the geometric series formula:
\[ P(x) = \frac{1 - x^4}{1 - x}. \]
The original expression becomes:
\[ \left( 1 + x + x^2 + x^3 \right)^5 = \left( \frac{1 - x^4}{1 - x} \right)^5. \]
Step 2: Expand. Rewrite the expression:
\[ \left( \frac{1 - x^4}{1 - x} \right)^5 = \left( 1 - x^4 \right)^5 \cdot \left( 1 - x \right)^{-5}. \]
Apply the binomial theorem:
\[ \left( 1 - x^4 \right)^5 = \sum_{m=0}^5 \binom{5}{m} (-1)^m x^{4m}, \]
\[ \left( 1 - x \right)^{-5} = \sum_{n=0}^\infty \binom{n+4}{4} x^n. \]
Step 3: Find coefficients. The coefficient \(a_{2k}\) comes from the product:
\[ \left( \sum_{m=0}^5 \binom{5}{m} (-1)^m x^{4m} \right) \cdot \left( \sum_{n=0}^\infty \binom{n+4}{4} x^n \right). \]
Equating powers, \(4m + n = 2k\), which implies:
\[ n = 2k - 4m. \]
Therefore, the coefficient of \(x^{2k}\) is:
\[ a_{2k} = \sum_{m=0}^{\lfloor k/2 \rfloor} \binom{5}{m} (-1)^m \binom{2k - 4m + 4}{4}. \]
Step 4: Calculate the sum. Substitute \(a_{2k}\) into the target sum:
\[ \sum_{k=0}^7 (-1)^k \cdot a_{2k} = \sum_{k=0}^7 (-1)^k \cdot \sum_{m=0}^{\lfloor k/2 \rfloor} \binom{5}{m} (-1)^m \binom{2k - 4m + 4}{4}. \]
Rearrange the summations:
\[ \sum_{k=0}^7 (-1)^k \cdot a_{2k} = \sum_{m=0}^5 \binom{5}{m} (-1)^m \sum_{k=0}^7 (-1)^k \binom{2k - 4m + 4}{4}. \]
The inner sum is zero due to the alternating sign, resulting in cancellation.
Conclusion: The result is:
\[ \boxed{0}. \]