Choose the factorised form of (a^2+b^2+2ab-c^2).

Correct answer: (a+b-c)(a+b+c)

The direct answer is A: $(a+b-c)(a+b+c)$. The first three terms form a perfect square: $a^2+2ab+b^2=(a+b)^2$. Therefore the expression becomes $(a+b)^2-c^2$. This is a difference of squares with $U=a+b$ and $V=c$. Applying $U^2-V^2=(U-V)(U+V)$ gives $[(a+b)-c][(a+b)+c]=(a+b-c)(a+b+c)$. Option A is correct. Expanding it as a difference of squares returns $(a+b)^2-c^2=a^2+2ab+b^2-c^2$. B is based on $(a-b)^2-c^2$ and would contain $-2ab$, not $+2ab$. C is a sum of squares, not the factorisation of a difference; it is not generally a product of the required two linear factors. D is based on $(a+c)^2-b^2$ and changes the roles of b and c, producing different terms. Memory cue: first combine three terms into a perfect square, then factor the remaining difference of squares.