Which of the following expressions can be factorised using the identity for the difference of two squares, \(p^2-q^2=(p+q)(p-q)\)?

Correct answer: \(9a^2-16b^2\)

The direct answer is A: \(9a^2-16b^2\). The difference-of-squares identity applies when two perfect-square terms are separated by a minus sign: \(P^2-Q^2=(P+Q)(P-Q)\). Here \(9a^2=(3a)^2\) and \(16b^2=(4b)^2\), so \(9a^2-16b^2=(3a+4b)(3a-4b)\). Expanding gives \(9a^2-12ab+12ab-16b^2=9a^2-16b^2\). B is a sum of squares, not a difference, so this identity does not apply. C is \((3a+4b)^2\) because its middle term is \(+24ab\). D is \((3a-4b)^2\) because its middle term is \(-24ab\). Those are perfect squares, not expressions of the required difference-of-squares form. A safe memory cue is to check two things: both terms must be squares, and the sign between them must be minus.