Which algebraic identity is suitable for factorising \(p^2-q^2\)?

Correct answer: \(a^2-b^2=(a-b)(a+b)\)

Direct answer: A. The expression \(p^2-q^2\) is a difference of two perfect squares: the squares are \(p^2\) and \(q^2\), separated by a minus sign. The identity \(a^2-b^2=(a-b)(a+b)\) therefore gives \(p^2-q^2=(p-q)(p+q)\). Expanding verifies it: \((p-q)(p+q)=p^2+pq-pq-q^2=p^2-q^2\). Option A is correct. Option B is the identity for the square of a sum and creates a positive middle term. Option C is the square of a difference and contains a middle term \(-2ab\), which is absent here. Option D is for the difference of cubes, not squares, and has three-factor-degree structure. Memory cue: two squared terms with a minus sign mean difference of squares: difference bracket times sum bracket.