What is the complete factorised form of \(3a^2−27b^2\)?

Correct answer: 3(a−3b)(a+3b)

Direct answer: A, 3(a−3b)(a+3b). First take out the common factor 3: 3a²−27b²=3(a²−9b²). Inside the brackets, 9b²=(3b)², so a²−9b² is a difference of squares. Apply u²−v²=(u−v)(u+v): a²−(3b)²=(a−3b)(a+3b). Hence the complete factorisation is 3(a−3b)(a+3b). Option B would expand to 3a²−3b², so it misses the factor 9 on b². Option C expands to 3a²+3ab−27ab−27b², which has unwanted middle terms. Option D represents a square and expands to 3a²−54ab+27b², with both a middle term and the wrong sign on the last term. Memory cue: remove the numerical common factor first, then look for a standard identity; do not stop at 3(a²−9b²).