Choose the factorised form of \(9x^2 − 1 + 6xy + y^2\).

Correct answer: (3x + y − 1)(3x + y + 1)

The direct answer is A. Rearrange the terms as \(9x^2+6xy+y^2−1\). The first three terms form a perfect square: \(9x^2+6xy+y^2=(3x+y)^2\), because \((3x+y)^2=9x^2+6xy+y^2\). Thus the expression becomes \((3x+y)^2−1^2\). Apply the difference-of-squares identity \(p^2−q^2=(p−q)(p+q)\), giving \((3x+y−1)(3x+y+1)\). Therefore A is correct. B would come from \((3x−y)^2−1\) and has \(−6xy\), not \(+6xy\). C does not preserve the correct leading and cross-term coefficients when expanded. D is only a square and expands to additional terms \(6x\), \(2y\) and a different constant, so it cannot equal the original expression. A common confusion is forgetting that the terms may be rearranged before identifying the identity. Memory cue: first make the perfect square, then subtract the remaining square.