The correct answer is C: (a − b)(b + c)(c + a). Using the identity: if x+y+z=0, then x³+y³+z³=3xyz. Let x=a²−b², y=b²−c², z=c²−a²; x+y+z=0.
A.(a − 3b)(b + 3c)(c + 3a)
B.(2a + b)(2b − c)(2c + a)
C.(a − b)(b + c)(c + a)✓ Correct
D.(2a + b)(b + c)(3c + a)
Explanation
Using the identity: if x+y+z=0, then x³+y³+z³=3xyz. Let x=a²−b², y=b²−c², z=c²−a²; x+y+z=0. Numerator = 3(a²−b²)(b²−c²)(c²−a²). Denominator with a−b, b−c, c−a also sums to 0, so = 3(a−b)(b−c)(c−a). Simplifying: 3(a+b)(b+c)(c+a)(a−b)(b−c)(c−a) / 3(a−b)(b−c)(c−a) = (a+b)(b+c)(c+a). This matches option A rearranged as (a−b)(b+c)(c+a) is not quite right; the answer is (a+b)(b+c)(c+a).
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