What is the factorised form of (x^4-2x^2y^2+y^4-z^2)?

Correct answer: ( (x^2-y^2-z)(x^2-y^2+z) )

The direct answer is C: \((x^2-y^2-z)(x^2-y^2+z)\). First recognize \(x^4-2x^2y^2+y^4\) as \((x^2-y^2)^2\), using \(A^2-2AB+B^2=(A-B)^2\) with \(A=x^2\), \(B=y^2\). The expression becomes \((x^2-y^2)^2-z^2\). Now apply the difference-of-squares identity \(P^2-Q^2=(P-Q)(P+Q)\), where \(P=x^2-y^2\) and \(Q=z\). This gives C. Option A uses \(x^2+y^2\), which changes the original middle sign. Option B has only linear x and y terms and cannot reproduce fourth-degree terms. Option C follows both identities exactly. Option D is a square, but the original expression is a difference of two squares and generally is not one square. Memory cue: first make the four-term part a square, then use difference of squares.