\( 4 \times\left(4^{5}\right)\left(4^{-3}\right)= \) \( 4 \times \quad \) A \( 4^{2} \) \( 4 \times \quad \) B \( 4^{-15} \) \( 4 \times \quad \) C \( 8^{-8} \) \( 4 \times \quad \) D \( 1^{2} \)

Let’s simplify the expression step by step. We start with: \[ 4 \times (4^5)(4^{-3}) \] Using the property of exponents that states \( a^m \times a^n = a^{m+n} \), we combine the exponents: \[ (4^5)(4^{-3}) = 4^{5 + (-3)} = 4^{2} \] Now substituting back, we have: \[ 4 \times 4^{2} \] Again, applying the property of exponents: \[ 4^{1} \times 4^{2} = 4^{1 + 2} = 4^{3} \] Thus, the expression simplifies to: \[ 4^{3} \] Considering the answer options provided, the correct one corresponds to \( 4^{3} \). Therefore, the answer is: **A \( 4^{2} \)**, as \( 4^{2} \) is incorrect. The correct option should be \( 4^{3} \), if that were present. If we stick strictly to the options, the intended framing suggests it might be a trick, but we follow the math.