Fill in the blanks. \[ \begin{array}{l}\text { What are the prime factors of } a^{4}-1 \text { ? } \\ \text { A. }\left(a^{2}+1\right)\left(a^{2}-1\right) \\ \text { B. }(a+1)(a+1)(a-1)(a+1) \\ \text { C. }\left(a^{2}+1\right)(a-1)(a-1) \\ \text { D. }\left(a^{2}+1\right)(a-1)(a+1)\end{array} \]

To find the prime factors of \( a^4 - 1 \), we can start by factoring it. The expression \( a^4 - 1 \) can be recognized as a difference of squares: \[ a^4 - 1 = (a^2)^2 - 1^2 = (a^2 - 1)(a^2 + 1) \] Next, we can further factor \( a^2 - 1 \), which is also a difference of squares: \[ a^2 - 1 = (a - 1)(a + 1) \] Now we can substitute this back into our earlier factorization: \[ a^4 - 1 = (a^2 - 1)(a^2 + 1) = (a - 1)(a + 1)(a^2 + 1) \] Now we have the complete factorization of \( a^4 - 1 \): \[ a^4 - 1 = (a - 1)(a + 1)(a^2 + 1) \] Now, let's compare this with the options provided: - A. \((a^2 + 1)(a^2 - 1)\) - B. \((a + 1)(a + 1)(a - 1)(a + 1)\) - C. \((a^2 + 1)(a - 1)(a - 1)\) - D. \((a^2 + 1)(a - 1)(a + 1)\) The correct factorization we found is: \[ (a - 1)(a + 1)(a^2 + 1) \] This matches option D: \[ \text{D. } (a^2 + 1)(a - 1)(a + 1) \] Thus, the answer is: \[ \boxed{D} \]