Explanation

Since x⁴−1 = (x−1)(x+1)(x²+1), divisibility requires P(1) = 0 and P(−1) = 0 for P(x) = x⁶+px⁵+qx⁴−x²−x−3. P(1) = p+q−4 = 0 gives p+q = 4, and P(−1) = q−p−2 = 0 gives q−p = 2. Solving yields p = 1, q = 3, so p²+q² = 1+9 = 10, option C.