What is the simplified form of ((x^6-64)/(x^3-8)), where (x^3≠8)?

Correct answer: x^3+8

The governing concept is the difference-of-squares identity, A^2−B^2=(A−B)(A+B). Rewrite the numerator as x^6−64=(x^3)^2−8^2. It therefore factors as (x^3−8)(x^3+8). Since the condition x^3≠8 guarantees that the denominator is non-zero, the common factor x^3−8 may be cancelled, leaving x^3+8. Thus option B is correct. Option A is only the cancelled denominator factor, option C does not follow from a valid factorization, and option D incorrectly adds the two square terms instead of applying the difference identity. The restriction remains important even after simplification because the original expression was undefined when x^3=8.